An informal investigation of a sum.

Consider this sum: $$S =\sum_{k=2}^{\infty}(k^{1/k} -1)$$ Does this converge? How does it behave as it diverges, if it diverges? If $k$ equaled $1$ we would get $0$ so we start at $k=2$. Very generally we can find that: $k^{1/k}$ will always be $1$ + a remainder. After subtracting $1$ from each term, each term will be a value less than $1$. Not too illuminating so far.

Let's raise the second k to a power t so we can examine how tweaking that variable affects the output $S_t = \sum_{k=2}^{\infty}(k^{1/k^t} -1)$. When $t=1$ it is the original and the partial sums seem to rise and rise, but is unclear of the final behavior.

Breaking $t$ out in parts, lets set $t = (1 + \frac{1}{x})$, $$S_x=\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1).$$

Numerically, these sums seem to converge when $x$ is finite, but its difficult to calculate when x gets large. Calculating some values, we find a pattern at last! We find $S_2 = 4$ plus a remainder less than $1$. Going further $S_3 = 9$ plus a remainder less than $1$, and $S_{100} = 10,000$ plus a remainder less than $1$. The output always seems to be the square plus a remainder! It seems true that $$\sum_{k=2}^{\infty}(k^{1/(k^{1+1/x})} -1) = \lfloor x^{2}\rfloor$$ when $x$ is the square root of any natural number greater than 1. Here are some examples: $S_4 = 16.238932773$, $S_{12} = 144.5937831$, $S_{\sqrt{1729}} = 1729.84841$, $S_{50000} = 2500000000.988421705$.

Does that remainder ever get higher than one? Let's get another perspective on the sum to see if we can calculate that remainder at the limit as ${x \to {\infty}}$.

Time to transform the sum so we can get a handle on why its the square plus a remainder, and try to get a handle on that remainder. Remembering that $t = (1 + 1/x)$, see that: $$k^{1/k^t}=e^{(\log k)/k^t} = \sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ Let's define $$S_t = \sum_{k=2}^{\infty}\left(\sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}\right) -1$$ Notice when $n=0$ we get $1$, so it cancels the $-1$ term resulting in: $$\sum_{k=2}^{\infty}\sum_{n = 1}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ What is effectively adding column by column instead of row by row, we get:$$\sum_{n=1}^{\infty}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}} = \sum_{n=1}^{\infty}\frac{1}{n!}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{k^{n t}}$$

Using the definition $$\zeta^{(n)}(t) = e^{i \pi n}\sum_{k=2}^{\infty}\frac{\log k ^{n}}{k^{t}}$$ valid for all $n\in\mathbb{R}, t\in\mathbb{C}$,

we get $$S_t = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n!}\zeta^{(n)}(nt)$$

Since we'll be looking at the Zeta function close to it's pole at $1$, let's expand $\zeta$ with the Laurent expansion of the Zeta function expanded around 1. $\gamma$ represents the Stieltjes constants.

$$\zeta(s) = \frac{1}{s-1} + \sum_{n=0}^{\infty}(-1)^{n}\frac{\gamma_n}{n!}(s-1)^{n}$$ Giving us $$\zeta^{'}(s) = -\frac1{(s-1)^2} -\gamma_1 +\gamma_2(s-1)-\frac12\gamma_3(s-1)^2\cdots$$ $$\zeta^{''}(s) = \frac2{(s-1)^3} +\gamma_2 -\gamma_3(s-1)+\frac12\gamma_4(s-1)^2\cdots$$ $$\zeta^{'''}(s) = -\frac6{(s-1)^4} -\gamma_3 +\gamma_4(s-1)-\frac12\gamma_5(s-1)^2...$$ Using these interpretations and setting $s = n(1+\frac1x)$ and examining what occurs when $x \to \infty$ allows us to see why the output of $S_x$ is always $x^2$ plus a constant less than 1. As before, we can add column by column instead of row by row for the summations of the Laurent series.

Examining the largest term when $n=1$, we have $$\frac1{((1+\frac1x)-1)^2} = x^2$$ $x^2$ obviously diverges, so already we can tell the original sum $S$ diverges. Let's keep going to see how the remainder term behaves as $x \to \infty$. When $n=2$ and greater, the terms such as $\frac{2}{(s-1)^3} = \frac2{(n-1)^3}$ turn into finite values. Summing all the terms in $S_t$, setting $t=1+\frac1x$ and letting x zoom off, we arrive at the limiting behavior of the sum $S_x$.

$$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{n=1}^{\infty}\frac{\gamma_n}{n!} + \sum_{k=3}^{\infty}\sum_{n=k}^{\infty}(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} = C,$$

and $$\sum_{n=1}^{\infty}\frac{\gamma_n}{n!} = \frac12 - \gamma_0,$$

and $$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{k=3}^{\infty}\sum_{n=k}^\infty(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} + \frac12 - \gamma_0 = C$$ .

$\gamma_n$ is bounded by $|\gamma_n| < \frac{n!}{2^{n+1}}$, which can show convergence, and numerically it stabilizes within the realm of testing.

$$C = 0.988549601142268750644\ldots$$

So it seems $$\sum_{k=2}^\infty (k^{1/(k^{1+1/x})} -1) = x^2 + C_x,$$ where $-0.028501\ldots < C_x < C$, $x\geq1$ , and $x \in \mathbb{R}$ , where $-0.028501\ldots$ is $C_x$ when $x=1$. $C_x$ is positive when roughly $x>~1.37$.

An informal investigation of a sum.

Consider this sum: $$S =\sum_{k=2}^{\infty}(k^{1/k} -1)$$ Does this converge? How does it behave as it diverges, if it diverges? If $k$ equaled $1$ we would get $0$ so we start at $k=2$. Very generally we can find that: $k^{1/k}$ will always be $1$ + a remainder. After subtracting $1$ from each term, each term will be a value less than $1$. Not too illuminating so far.

Let's raise the second k to a power t so we can examine how tweaking that variable affects the output $S_t = \sum_{k=2}^{\infty}(k^{1/k^t} -1)$. When $t=1$ it is the original and the partial sums seem to rise and rise, but is unclear of the final behavior.

Breaking $t$ out in parts, lets set $t = (1 + \frac{1}{x})$, $$S_x=\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1).$$

Numerically, these sums seem to converge when $x$ is finite, but its difficult to calculate when x gets large. Calculating some values, we find a pattern at last! We find $S_2 = 4$ plus a remainder less than $1$. Going further $S_3 = 9$ plus a remainder less than $1$, and $S_{100} = 10,000$ plus a remainder less than $1$. The output always seems to be the square plus a remainder! It seems true that $$\sum_{k=2}^{\infty}(k^{1/(k^{1+1/x})} -1) = \lfloor x^{2}\rfloor$$ when $x$ is the square root of any natural number greater than 1. Here are some examples: $S_4 = 16.238932773$, $S_{12} = 144.5937831$, $S_{\sqrt{1729}} = 1729.84841$, $S_{50000} = 2500000000.988421705$.

Does that remainder ever get higher than one? Let's get another perspective on the sum to see if we can calculate that remainder at the limit as ${x \to {\infty}}$.

Time to transform the sum so we can get a handle on why its the square plus a remainder, and try to get a handle on that remainder. Remembering that $t = (1 + 1/x)$, see that: $$k^{1/k^t}=e^{(\log k)/k^t} = \sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ Let's define $$S_t = \sum_{k=2}^{\infty}\left(\sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}\right) -1$$ Notice when $n=0$ we get $1$, so it cancels the $-1$ term resulting in: $$\sum_{k=2}^{\infty}\sum_{n = 1}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ What is effectively adding column by column instead of row by row, we get:$$\sum_{n=1}^{\infty}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}} = \sum_{n=1}^{\infty}\frac{1}{n!}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{k^{n t}}$$

Using the definition $$\zeta^{(n)}(t) = e^{i \pi n}\sum_{k=2}^{\infty}\frac{\log k ^{n}}{k^{t}}$$ valid for all $n\in\mathbb{R}, t\in\mathbb{C}$,

we get $$S_t = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n!}\zeta^{(n)}(nt)$$

Since we'll be looking at the Zeta function close to it's pole at $1$, let's expand $\zeta$ with the Laurent expansion of the Zeta function expanded around 1. $\gamma$ represents the Stieltjes constants.

$$\zeta(s) = \frac{1}{s-1} + \sum_{n=0}^{\infty}(-1)^{n}\frac{\gamma_n}{n!}(s-1)^{n}$$ Giving us $$\zeta^{'}(s) = -\frac1{(s-1)^2} -\gamma_1 +\gamma_2(s-1)-\frac12\gamma_3(s-1)^2\cdots$$ $$\zeta^{''}(s) = \frac2{(s-1)^3} +\gamma_2 -\gamma_3(s-1)+\frac12\gamma_4(s-1)^2\cdots$$ $$\zeta^{'''}(s) = -\frac6{(s-1)^4} -\gamma_3 +\gamma_4(s-1)-\frac12\gamma_5(s-1)^2...$$ Using these interpretations and setting $s = n(1+\frac1x)$ and examining what occurs when $x \to \infty$ allows us to see why the output of $S_x$ is always $x^2$ plus a constant less than 1. As before, we can add column by column instead of row by row for the summations of the Laurent series.

Examining the largest term when $n=1$, we have $$\frac1{((1+\frac1x)-1)^2} = x^2$$ $x^2$ obviously diverges, so already we can tell the original sum $S$ diverges. Let's keep going to see how the remainder term behaves as $x \to \infty$. When $n=2$ and greater, the terms such as $\frac{2}{(s-1)^3} = \frac2{(n-1)^3}$ turn into finite values. Summing all the terms in $S_t$, setting $t=1+\frac1x$ and letting x zoom off, we arrive at the limiting behavior of the sum $S_x$.

$$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{n=1}^{\infty}\frac{\gamma_n}{n!} + \sum_{k=3}^{\infty}\sum_{n=k}^{\infty}(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} = C,$$

and $$\sum_{n=1}^{\infty}\frac{\gamma_n}{n!} = \frac12 - \gamma_0,$$

and $$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{k=3}^{\infty}\sum_{n=k}^\infty(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} + \frac12 - \gamma_0 = C$$ .

$\gamma_n$ is bounded by $|\gamma_n| < \frac{n!}{2^{n+1}}$, which can show convergence, and numerically it stabilizes within the realm of testing.

$$C = 0.988549601142268750644\ldots$$

So it seems $$\sum_{k=2}^\infty (k^{1/(k^{1+1/x})} -1) = x^2 + C_x,$$ where $-0.028501\ldots < C_x < C$, $x\geq1$ , and $x \in \mathbb{R}$ , where $-0.028501\ldots$ is $C_x$ when $x=1$. $C_x$ is positive when roughly $x>~1.37$.

Edit: Hi everyone, I didn't know the strict format of this site. Please see the selected answer as a beautiful solution to my first question, and please consider the rest of this post as exposition of how the sum diverges.

An informal investigation of a sum.

Consider this sum: $$S =\sum_{k=2}^{\infty}(k^{1/k} -1)$$ Does this converge? How does it behave as it diverges, if it diverges? If $k$ equaled $1$ we would get $0$ so we start at $k=2$. Very generally we can find that: $k^{1/k}$ will always be $1$ + a remainder. After subtracting $1$ from each term, each term will be a value less than $1$. Not too illuminating so far.

Let's raise the second k to a power t so we can examine how tweaking that variable affects the output $S_t = \sum_{k=2}^{\infty}(k^{1/k^t} -1)$. When $t=1$ it is the original and the partial sums seem to rise and rise, but is unclear of the final behavior.

Breaking $t$ out in parts, lets set $t = (1 + \frac{1}{x})$, $$S_x=\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1).$$

Numerically, these sums seem to converge when $x$ is finite, but its difficult to calculate when x gets large. Calculating some values, we find a pattern at last! We find $S_2 = 4$ plus a remainder less than $1$. Going further $S_3 = 9$ plus a remainder less than $1$, and $S_{100} = 10,000$ plus a remainder less than $1$. The output always seems to be the square plus a remainder! It seems true that $$\sum_{k=2}^{\infty}(k^{1/(k^{1+1/x})} -1) = \lfloor x^{2}\rfloor$$ when $x^{2}$ is a natural number greater than 1. Here are some examples: $S_4 = 16.238932773$, $S_{12} = 144.5937831$, $S_{\sqrt{1729}} = 1729.84841$, $S_{50000} = 2500000000.988421705$.

Does that remainder ever get higher than one? Let's get another perspective on the sum to see if we can calculate that remainder at the limit as ${x \to {\infty}}$.

Time to transform the sum so we can get a handle on why its the square plus a remainder, and try to get a handle on that remainder. Remembering that $t = (1 + 1/x)$, see that: $$k^{1/k^t}=e^{(\log k)/k^t} = \sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ Let's define $$S_t = \sum_{k=2}^{\infty}\left(\sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}\right) -1$$ Notice when $n=0$ we get $1$, so it cancels the $-1$ term resulting in: $$\sum_{k=2}^{\infty}\sum_{n = 1}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ What is effectively adding column by column instead of row by row, we get:$$\sum_{n=1}^{\infty}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}} = \sum_{n=1}^{\infty}\frac{1}{n!}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{k^{n t}}$$

Using the definition $$\zeta^{(n)}(t) = e^{i \pi n}\sum_{k=2}^{\infty}\frac{\log k ^{n}}{k^{t}}$$ valid for all $n\in\mathbb{R}, t\in\mathbb{C}$,

we get $$S_t = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n!}\zeta^{(n)}(nt)$$

Since we'll be looking at the Zeta function close to it's pole at $1$, let's expand $\zeta$ with the Laurent expansion of the Zeta function expanded around 1. $\gamma$ represents the Stieltjes constants.

$$\zeta(s) = \frac{1}{s-1} + \sum_{n=0}^{\infty}(-1)^{n}\frac{\gamma_n}{n!}(s-1)^{n}$$ Giving us $$\zeta^{'}(s) = -\frac1{(s-1)^2} -\gamma_1 +\gamma_2(s-1)-\frac12\gamma_3(s-1)^2\cdots$$ $$\zeta^{''}(s) = \frac2{(s-1)^3} +\gamma_2 -\gamma_3(s-1)+\frac12\gamma_4(s-1)^2\cdots$$ $$\zeta^{'''}(s) = -\frac6{(s-1)^4} -\gamma_3 +\gamma_4(s-1)-\frac12\gamma_5(s-1)^2...$$ Using these interpretations and setting $s = n(1+\frac1x)$ and examining what occurs when $x \to \infty$ allows us to see why the output of $S_x$ is always $x^2$ plus a constant less than 1. As before, we can add column by column instead of row by row for the summations of the Laurent series.

Examining the largest term when $n=1$, we have $$\frac1{((1+\frac1x)-1)^2} = x^2$$ $x^2$ obviously diverges, so already we can tell the original sum $S$ diverges. Let's keep going to see how the remainder term behaves as $x \to \infty$. When $n=2$ and greater, the terms such as $\frac{2}{(s-1)^3} = \frac2{(n-1)^3}$ turn into finite values. Summing all the terms in $S_t$, setting $t=1+\frac1x$ and letting x zoom off, we arrive at the limiting behavior of the sum $S_x$.

$$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{n=1}^{\infty}\frac{\gamma_n}{n!} + \sum_{k=3}^{\infty}\sum_{n=k}^{\infty}(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} = C,$$

and $$\sum_{n=1}^{\infty}\frac{\gamma_n}{n!} = \frac12 - \gamma_0,$$

and $$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{k=3}^{\infty}\sum_{n=k}^\infty(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} + \frac12 - \gamma_0 = C$$ .

$\gamma_n$ is bounded by $|\gamma_n| < \frac{n!}{2^{n+1}}$, which can show convergence, and numerically it stabilizes within the realm of testing.

$$C = 0.988549601142268750644\ldots$$

So it seems $$\sum_{k=2}^\infty (k^{1/(k^{1+1/x})} -1) = x^2 + C_x,$$ where $-0.028501\ldots < C_x < C$, $x\geq1$ , and $x \in \mathbb{R}$ , where $-0.028501\ldots$ is $C_x$ when $x=1$. $C_x$ is positive when roughly $x>~1.37$.

An informal investigation of a sum.

Consider this sum: $$S =\sum_{k=2}^{\infty}(k^{1/k} -1)$$ Does this converge? How does it behave as it diverges, if it diverges? If $k$ equaled $1$ we would get $0$ so we start at $k=2$. Very generally we can find that: $k^{1/k}$ will always be $1$ + a remainder. After subtracting $1$ from each term, each term will be a value less than $1$. Not too illuminating so far.

Let's raise the second k to a power t so we can examine how tweaking that variable affects the output $S_t = \sum_{k=2}^{\infty}(k^{1/k^t} -1)$. When $t=1$ it is the original and the partial sums seem to rise and rise, but is unclear of the final behavior.

Breaking $t$ out in parts, lets set $t = (1 + \frac{1}{x})$, $$S_x=\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1).$$

Numerically, these sums seem to converge when $x$ is finite, but its difficult to calculate when x gets large. Calculating some values, we find a pattern at last! We find $S_2 = 4$ plus a remainder less than $1$. Going further $S_3 = 9$ plus a remainder less than $1$, and $S_{100} = 10,000$ plus a remainder less than $1$. The output always seems to be the square plus a remainder! It seems true that $$\sum_{k=2}^{\infty}(k^{1/(k^{1+1/x})} -1) = \lfloor x^{2}\rfloor$$ when $x$ is the square root of any natural number greater than 1. Here are some examples: $S_4 = 16.238932773$, $S_{12} = 144.5937831$, $S_{\sqrt{1729}} = 1729.84841$, $S_{50000} = 2500000000.988421705$.

Does that remainder ever get higher than one? Let's get another perspective on the sum to see if we can calculate that remainder at the limit as ${x \to {\infty}}$.

Time to transform the sum so we can get a handle on why its the square plus a remainder, and try to get a handle on that remainder. Remembering that $t = (1 + 1/x)$, see that: $$k^{1/k^t}=e^{(\log k)/k^t} = \sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ Let's define $$S_t = \sum_{k=2}^{\infty}\left(\sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}\right) -1$$ Notice when $n=0$ we get $1$, so it cancels the $-1$ term resulting in: $$\sum_{k=2}^{\infty}\sum_{n = 1}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ What is effectively adding column by column instead of row by row, we get:$$\sum_{n=1}^{\infty}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}} = \sum_{n=1}^{\infty}\frac{1}{n!}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{k^{n t}}$$

Using the definition $$\zeta^{(n)}(t) = e^{i \pi n}\sum_{k=2}^{\infty}\frac{\log k ^{n}}{k^{t}}$$ valid for all $n\in\mathbb{R}, t\in\mathbb{C}$,

we get $$S_t = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n!}\zeta^{(n)}(nt)$$

Since we'll be looking at the Zeta function close to it's pole at $1$, let's expand $\zeta$ with the Laurent expansion of the Zeta function expanded around 1. $\gamma$ represents the Stieltjes constants.

$$\zeta(s) = \frac{1}{s-1} + \sum_{n=0}^{\infty}(-1)^{n}\frac{\gamma_n}{n!}(s-1)^{n}$$ Giving us $$\zeta^{'}(s) = -\frac1{(s-1)^2} -\gamma_1 +\gamma_2(s-1)-\frac12\gamma_3(s-1)^2\cdots$$ $$\zeta^{''}(s) = \frac2{(s-1)^3} +\gamma_2 -\gamma_3(s-1)+\frac12\gamma_4(s-1)^2\cdots$$ $$\zeta^{'''}(s) = -\frac6{(s-1)^4} -\gamma_3 +\gamma_4(s-1)-\frac12\gamma_5(s-1)^2...$$ Using these interpretations and setting $s = n(1+\frac1x)$ and examining what occurs when $x \to \infty$ allows us to see why the output of $S_x$ is always $x^2$ plus a constant less than 1. As before, we can add column by column instead of row by row for the summations of the Laurent series.

Examining the largest term when $n=1$, we have $$\frac1{((1+\frac1x)-1)^2} = x^2$$ $x^2$ obviously diverges, so already we can tell the original sum $S$ diverges. Let's keep going to see how the remainder term behaves as $x \to \infty$. When $n=2$ and greater, the terms such as $\frac{2}{(s-1)^3} = \frac2{(n-1)^3}$ turn into finite values. Summing all the terms in $S_t$, setting $t=1+\frac1x$ and letting x zoom off, we arrive at the limiting behavior of the sum $S_x$.

$$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{n=1}^{\infty}\frac{\gamma_n}{n!} + \sum_{k=3}^{\infty}\sum_{n=k}^{\infty}(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} = C,$$

and $$\sum_{n=1}^{\infty}\frac{\gamma_n}{n!} = \frac12 - \gamma_0,$$

and $$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{k=3}^{\infty}\sum_{n=k}^\infty(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} + \frac12 - \gamma_0 = C$$ .

$\gamma_n$ is bounded by $|\gamma_n| < \frac{n!}{2^{n+1}}$, which can show convergence, and numerically it stabilizes within the realm of testing.

$$C = 0.988549601142268750644\ldots$$

So it seems $$\sum_{k=2}^\infty (k^{1/(k^{1+1/x})} -1) = x^2 + C_x,$$ where $-0.028501\ldots < C_x < C$, $x\geq1$ , and $x \in \mathbb{R}$ , where $-0.028501\ldots$ is $C_x$ when $x=1$. $C_x$ is positive when roughly $x>~1.37$.

An informal investigation of a sum.

Consider this sum: $$S =\sum_{k=2}^{\infty}(k^{1/k} -1)$$ How does it behave as it diverges, if it diverges? If $k$ equaled $1$ we would get $0$ so we start at $k=2$. Very generally we can find that: $k^{1/k}$ will always be $1$ + a remainder. After subtracting $1$ from each term, each term will be a value less than $1$. Not too illuminating so far.

Let's raise the second k to a power t so we can examine how tweaking that variable affects the output $S_t = \sum_{k=2}^{\infty}(k^{1/k^t} -1)$. When $t=1$ it is the original and the partial sums seem to rise and rise, but is unclear of the final behavior.

Breaking $t$ out in parts, lets set $t = (1 + \frac{1}{x})$, $$S_x=\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1).$$

Numerically, these sums seem to converge when $x$ is finite, but its difficult to calculate when x gets large. Calculating some values, we find a pattern at last! We find $S_2 = 4$ plus a remainder less than $1$. Going further $S_3 = 9$ plus a remainder less than $1$, and $S_{100} = 10,000$ plus a remainder less than $1$. The output always seems to be the square plus a remainder! It seems true that $$\sum_{k=2}^{\infty}(k^{1/(k^{1+1/x})} -1) = \lfloor x^{2}\rfloor$$ when $x^{2}$ is a natural number greater than 1. Here are some examples: $S_4 = 16.238932773$, $S_{12} = 144.5937831$, $S_{\sqrt{1729}} = 1729.84841$, $S_{50000} = 2500000000.988421705$.

Does that remainder ever get higher than one? Let's get another perspective on the sum to see if we can calculate that remainder at the limit as ${x \to {\infty}}$.

Time to transform the sum so we can get a handle on why its the square plus a remainder, and try to get a handle on that remainder. Remembering that $t = (1 + 1/x)$, see that: $$k^{1/k^t}=e^{(\log k)/k^t} = \sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ Let's define $$S_t = \sum_{k=2}^{\infty}\left(\sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}\right) -1$$ Notice when $n=0$ we get $1$, so it cancels the $-1$ term resulting in: $$\sum_{k=2}^{\infty}\sum_{n = 1}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ What is effectively adding column by column instead of row by row, we get:$$\sum_{n=1}^{\infty}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}} = \sum_{n=1}^{\infty}\frac{1}{n!}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{k^{n t}}$$

Using the definition $$\zeta^{(n)}(t) = e^{i \pi n}\sum_{k=2}^{\infty}\frac{\log k ^{n}}{k^{t}}$$ valid for all $n\in\mathbb{R}, t\in\mathbb{C}$,

we get $$S_t = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n!}\zeta^{(n)}(nt)$$

Since we'll be looking at the Zeta function close to it's pole at $1$, let's expand $\zeta$ with the Laurent expansion of the Zeta function expanded around 1. $\gamma$ represents the Stieltjes constants.

$$\zeta(s) = \frac{1}{s-1} + \sum_{n=0}^{\infty}(-1)^{n}\frac{\gamma_n}{n!}(s-1)^{n}$$ Giving us $$\zeta^{'}(s) = -\frac1{(s-1)^2} -\gamma_1 +\gamma_2(s-1)-\frac12\gamma_3(s-1)^2\cdots$$ $$\zeta^{''}(s) = \frac2{(s-1)^3} +\gamma_2 -\gamma_3(s-1)+\frac12\gamma_4(s-1)^2\cdots$$ $$\zeta^{'''}(s) = -\frac6{(s-1)^4} -\gamma_3 +\gamma_4(s-1)-\frac12\gamma_5(s-1)^2...$$ Using these interpretations and setting $s = n(1+\frac1x)$ and examining what occurs when $x \to \infty$ allows us to see why the output of $S_x$ is always $x^2$ plus a constant less than 1. As before, we can add column by column instead of row by row for the summations of the Laurent series.

Examining the largest term when $n=1$, we have $$\frac1{((1+\frac1x)-1)^2} = x^2$$ $x^2$ obviously diverges, so already we can tell the original sum $S$ diverges. Let's keep going to see how the remainder term behaves as $x \to \infty$. When $n=2$ and greater, the terms such as $\frac{2}{(s-1)^3} = \frac2{(n-1)^3}$ turn into finite values. Summing all the terms in $S_t$, setting $t=1+\frac1x$ and letting x zoom off, we arrive at the limiting behavior of the sum $S_x$.

$$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{n=1}^{\infty}\frac{\gamma_n}{n!} + \sum_{k=3}^{\infty}\sum_{n=k}^{\infty}(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} = C,$$

and $$\sum_{n=1}^{\infty}\frac{\gamma_n}{n!} = \frac12 - \gamma_0,$$

and $$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{k=3}^{\infty}\sum_{n=k}^\infty(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} + \frac12 - \gamma_0 = C$$ .

$\gamma_n$ is bounded by $|\gamma_n| < \frac{n!}{2^{n+1}}$, which can show convergence, and numerically it stabilizes within the realm of testing.

$$C = 0.988549601142268750644\ldots$$

So it seems $$\sum_{k=2}^\infty (k^{1/(k^{1+1/x})} -1) = x^2 + C_x,$$ where $-0.028501\ldots < C_x < C$, $x\geq1$ , and $x \in \mathbb{R}$ , where $-0.028501\ldots$ is $C_x$ when $x=1$. $C_x$ is positive when roughly $x>~1.37$.

An informal investigation of a sum.

Consider this sum: $$S =\sum_{k=2}^{\infty}(k^{1/k} -1)$$ Does this converge? How does it behave as it diverges, if it diverges? If $k$ equaled $1$ we would get $0$ so we start at $k=2$. Very generally we can find that: $k^{1/k}$ will always be $1$ + a remainder. After subtracting $1$ from each term, each term will be a value less than $1$. Not too illuminating so far.

Let's raise the second k to a power t so we can examine how tweaking that variable affects the output $S_t = \sum_{k=2}^{\infty}(k^{1/k^t} -1)$. When $t=1$ it is the original and the partial sums seem to rise and rise, but is unclear of the final behavior.

Breaking $t$ out in parts, lets set $t = (1 + \frac{1}{x})$, $$S_x=\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1).$$

Numerically, these sums seem to converge when $x$ is finite, but its difficult to calculate when x gets large. Calculating some values, we find a pattern at last! We find $S_2 = 4$ plus a remainder less than $1$. Going further $S_3 = 9$ plus a remainder less than $1$, and $S_{100} = 10,000$ plus a remainder less than $1$. The output always seems to be the square plus a remainder! It seems true that $$\sum_{k=2}^{\infty}(k^{1/(k^{1+1/x})} -1) = \lfloor x^{2}\rfloor$$ when $x^{2}$ is a natural number greater than 1. Here are some examples: $S_4 = 16.238932773$, $S_{12} = 144.5937831$, $S_{\sqrt{1729}} = 1729.84841$, $S_{50000} = 2500000000.988421705$.

Does that remainder ever get higher than one? Let's get another perspective on the sum to see if we can calculate that remainder at the limit as ${x \to {\infty}}$.

Time to transform the sum so we can get a handle on why its the square plus a remainder, and try to get a handle on that remainder. Remembering that $t = (1 + 1/x)$, see that: $$k^{1/k^t}=e^{(\log k)/k^t} = \sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ Let's define $$S_t = \sum_{k=2}^{\infty}\left(\sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}\right) -1$$ Notice when $n=0$ we get $1$, so it cancels the $-1$ term resulting in: $$\sum_{k=2}^{\infty}\sum_{n = 1}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ What is effectively adding column by column instead of row by row, we get:$$\sum_{n=1}^{\infty}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}} = \sum_{n=1}^{\infty}\frac{1}{n!}\sum_{k=2}^{\infty} \frac{\left( \log k \right)^n}{k^{n t}}$$

Using the definition $$\zeta^{(n)}(t) = e^{i \pi n}\sum_{k=2}^{\infty}\frac{\log k ^{n}}{k^{t}}$$ valid for all $n\in\mathbb{R}, t\in\mathbb{C}$,

we get $$S_t = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n!}\zeta^{(n)}(nt)$$

Since we'll be looking at the Zeta function close to it's pole at $1$, let's expand $\zeta$ with the Laurent expansion of the Zeta function expanded around 1. $\gamma$ represents the Stieltjes constants.

$$\zeta(s) = \frac{1}{s-1} + \sum_{n=0}^{\infty}(-1)^{n}\frac{\gamma_n}{n!}(s-1)^{n}$$ Giving us $$\zeta^{'}(s) = -\frac1{(s-1)^2} -\gamma_1 +\gamma_2(s-1)-\frac12\gamma_3(s-1)^2\cdots$$ $$\zeta^{''}(s) = \frac2{(s-1)^3} +\gamma_2 -\gamma_3(s-1)+\frac12\gamma_4(s-1)^2\cdots$$ $$\zeta^{'''}(s) = -\frac6{(s-1)^4} -\gamma_3 +\gamma_4(s-1)-\frac12\gamma_5(s-1)^2...$$ Using these interpretations and setting $s = n(1+\frac1x)$ and examining what occurs when $x \to \infty$ allows us to see why the output of $S_x$ is always $x^2$ plus a constant less than 1. As before, we can add column by column instead of row by row for the summations of the Laurent series.

Examining the largest term when $n=1$, we have $$\frac1{((1+\frac1x)-1)^2} = x^2$$ $x^2$ obviously diverges, so already we can tell the original sum $S$ diverges. Let's keep going to see how the remainder term behaves as $x \to \infty$. When $n=2$ and greater, the terms such as $\frac{2}{(s-1)^3} = \frac2{(n-1)^3}$ turn into finite values. Summing all the terms in $S_t$, setting $t=1+\frac1x$ and letting x zoom off, we arrive at the limiting behavior of the sum $S_x$.

$$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{n=1}^{\infty}\frac{\gamma_n}{n!} + \sum_{k=3}^{\infty}\sum_{n=k}^{\infty}(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} = C,$$

and $$\sum_{n=1}^{\infty}\frac{\gamma_n}{n!} = \frac12 - \gamma_0,$$

and $$\lim_{x \to \infty}\Big|\Big(\sum_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum_{k=3}^{\infty}\sum_{n=k}^\infty(-1)^k\frac{\gamma_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} + \frac12 - \gamma_0 = C$$ .

$\gamma_n$ is bounded by $|\gamma_n| < \frac{n!}{2^{n+1}}$, which can show convergence, and numerically it stabilizes within the realm of testing.

$$C = 0.988549601142268750644\ldots$$

So it seems $$\sum_{k=2}^\infty (k^{1/(k^{1+1/x})} -1) = x^2 + C_x,$$ where $-0.028501\ldots < C_x < C$, $x\geq1$ , and $x \in \mathbb{R}$ , where $-0.028501\ldots$ is $C_x$ when $x=1$. $C_x$ is positive when roughly $x>~1.37$.