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We can consider the generalized Harmonic numbers $$H_{n,m} := \sum_{k=1}^{n} \frac{1}{k^{m}} $$ as a partial version of the Riemann zeta function, because $$\lim_{n \to \infty} H_{n,m} = \zeta(m). $$

We could also define the "partial Sophomore's Dream function" $$S_{r} := \sum_{q=1}^{r}\frac{1}{q^{q}} ,$$ as we have $$\lim_{r \to \infty} S_{r} := S= \int_{0}^{1}x^{-x}dx ,$$ where the integral on the right is equal to the first Sophomore's Dream constant.

Question: while studying the series $$A := \sum_{r=1}^{\infty}(S-S_{r}) \approx 0.3371877158, $$ I was wondering whether the function $S_{r}$ has already been studied, and if someone has coined a name for it that is hopefully less awkward than mine. I am looking for alternative representations of $S_{r}$ that might help finding a closed form of $A$.

Note: this question was previously asked on MSE.

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2 Answers 2

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We have that $S=\sum_{n=1}^\infty \frac 1 {n^n}$ and $S_r=\sum_{n=1}^r \frac 1 {n^n}$. Thus,

$A=\sum_{r=1}^\infty \sum_{n=r+1}^\infty \frac 1 {n^n} = \sum_{n=2}^\infty \sum_{r=1}^{n-1} \frac 1 {n^n}$

But then, $A=\sum_{n=2}^\infty \frac {n-1} {n^n}$. To evaluate this, consider the integral $\int_0^1 t^{-tx} dt = \sum_{n=1}^\infty \frac {x^{n-1}} {n^n}$. Differentiating with respect to $x$, we get $\sum_{n=2}^\infty \frac {(n-1)x^{n-2}} {n^n} = \int_0^1 -t\ln t \cdot t^{-xt} dt$. And substituting $x=1$ finally gives

$A=\int_0^1 -\ln t \cdot t^{1-t} dt$.

Note that a WolframAlpha calculation gives 0.3371877 as the value of the integral, in excellent agreement with the value you obtained.

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  • $\begingroup$ Thank you! Minor nitpick: is there maybe a small typo in the second equality for $A$? There the inner sum goes from $r=1$ to $r=n-1$, but there's no $r$ in the summands $\endgroup$ Commented Feb 15, 2022 at 10:16
  • $\begingroup$ @Max Muller Yes, that is correct. What I did there is just reversing the two summations, so the summand itself stays unchanged. And exactly because there is no $r$, we can simplify to the single sum. $\endgroup$
    – FusRoDah
    Commented Feb 15, 2022 at 16:32
  • $\begingroup$ Ah I see! Interesting result $\endgroup$ Commented Feb 15, 2022 at 16:47
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I don't know about an alternative name, but your query for "alternative representations of $S_r$" can be answered by the identity $$\sum_{q=1}^r \frac{1}{q^q}=\int_0^1 \frac{\Gamma(r,-x\ln x)}{x^x\, \Gamma(r)}dx.$$ Since $\lim_{r\rightarrow\infty} \Gamma(r,a)/\Gamma(r)=1$, in that limit you recover the sophomore's dream.

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