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I've found through evidence and have conjectured on a math publication that:

$$\Big\lfloor\int_1^\infty (k^{1/(k^{1+1/\sqrt{x}})} - 1)dk\Big\rfloor = \Big\lfloor\sum_{k=1}^{\infty}k^{1/(k^{1+1/\sqrt{x}})} -1\Big\rfloor = x $$

where $ x \in \mathbb{N}, x>1$.

It is very hard to compute these values. Repeated Shanks transformations and Richardson's Extrapolation will be required to compute, or using Pari GP techniques. Before you post a counter example below 10^7 for the sum, please check your precision.

Proving this has proved extremely difficult.

My question is, does anyone have any suggestions of how to prove this? The only information I have is that this is true from all tests for $x$ less than 10^7 and we're still running tests for the sums.

They aren't equal without the floor function, and each equal $x + C$, where $C$ is a constant less than 1, and $C$ is different for the integral and sum. As $x$ tends to infinity, $C$ tends to 1.

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  • $\begingroup$ Your subject line said that you were asking for help, and your last paragraph asked people not to get angry. I think neither of these adds to the problem—a question might or might not be well received, but I would say this community almost never gets angry at someone asking a question in good faith. I have edited them out, but, if you are sure that they add to the problem, please feel free to edit them back in. (Introducing $x$, and most of your last paragraph, seem redundant; it seems to say no more than the equality already asserted. If you agree, then you might edit that, too.) $\endgroup$
    – LSpice
    Jan 26 at 21:08
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    $\begingroup$ Thank you I appreciate your edits. The last sentence was to mostly state that C tends to 1, as x goes to infinity, not necessarily to state again that there would be a remainder less than 1. But, I am open to all suggestions and edits frankly. I accidentally posted this on math.se and thats the wrong place for conjecture so I'm a little shy. $\endgroup$
    – user475930
    Jan 26 at 21:13
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    $\begingroup$ for $x=1$ I obtain 1.0387 for the integral and 0.971499 for the sum, so their integer part is not equal --- you say that this is an issue with numerical precision? $\endgroup$ Jan 26 at 21:19
  • $\begingroup$ My apologies, I forgot to include the stipulation that x is greater than 1. I transcribed this from another document. $\endgroup$
    – user475930
    Jan 26 at 21:40

1 Answer 1

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First of all, $$k^{1/k^t}=e^{(\log k)/k^t} = \sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ and therefore $$\intop_{1}^{\infty} \left( k^{1/k^t} - 1 \right) d k = \sum_{n = 1}^{\infty} \frac{1}{n!} \intop_{1}^{\infty} \left( \log k \right)^n k^{- n t} d k =$$ $$\sum_{n = 1}^{\infty} \frac{1}{n!} \intop_{0}^{\infty} y^n e^{(1 - n t) y} d y = \sum_{n = 1}^{\infty} \frac{1}{(n t - 1)^{n + 1}}.$$ If $t = 1 + 1/\sqrt{x}$ then the main contribution to this sum is the term $n = 1$ which gives a contribution of $x$, and so the sum is equal to $$x + \sum_{n = 2}^{\infty} \frac{1}{(n t - 1)^{n + 1}}.$$ In order for the floor of this value to be equal to $x$, then we must have $\sum_{n = 2}^{\infty} \frac{1}{(n t - 1)^{n + 1}} < 1$ for each $t > 1$, or equivalently $\sum_{n = 2}^{\infty} \frac{1}{(n - 1)^{n + 1}} \leq 1$. However, this is false: indeed, the second term alone is equal to $1$, so for large enough $x$ the floor of the integral is at least $x + 1$. What is true is that the integral is always strictly less than $x + 2$, since $$\sum_{n = 3}^{\infty} \frac{1}{(n - 1)^{n + 1}} < \sum_{n = 3}^{\infty} \frac{1}{2^{n + 1}} = \frac{1}{8}.$$


Presumably, the sum can be treated in a similar manner, where the main term (which should be about $x$) would be $\sum_{k = 1}^{\infty} \frac{\log k}{k^{1 + 1/\sqrt{x}}}$, and the other terms contributing at most a constant.

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  • $\begingroup$ By the way, Wolfram Alpha says that already for $t = 1.01$ we have $\sum_{n = 2}^{\infty} \frac{1}{(n t - 1)^{n + 1}} > 1$, that is already for $x = 10^4$ the floor of the integral is equal to $x + 1$. $\endgroup$
    – Random
    Jan 26 at 21:41
  • $\begingroup$ @TheHoyt, whether or not begging is appropriate, you made 5 requests in about 20 minutes. @‍Random may or may not help, but at least it is appropriate to give them the courtesy of a bit of time to consider and respond, if they choose to do so. $\endgroup$
    – LSpice
    Jan 26 at 22:26
  • $\begingroup$ Thank you for not spoiling the fun :D That was supremely fun. $\endgroup$
    – user475930
    Feb 3 at 3:00
  • $\begingroup$ The computed limit for the sum is x + 0.9885435... and is terribly hard to compute due to needing the Stieltje constants. Going to make a new question on here with my 'proof' of the sum. $\endgroup$
    – user475930
    Feb 7 at 15:53
  • $\begingroup$ That series1/(n-1)^n+1 is in the estimation in the sum as well, but there are an infinite amount of correction terms that bring it down below one, mainly because Stieltjes(1) is negative. $\endgroup$
    – user475930
    Feb 7 at 19:06

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