Evaluating an infinite series in terms of the Stieltjes Constants

Motivated by this Math Stack Exchange problem the goal is to evaluate the sum

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}\log^k n}{n}$

for any non-negative integer $k$. From looking at this, it is very reasonable to expect that the answer should be a linear combination of the Stieltjes Constants, $\gamma_n$ for $n\leq k$, since by definition

$\displaystyle \gamma_{n}:=\lim_{m\rightarrow\infty}\left(\sum_{j=1}^{m}\frac{\log^{n}j}{j}-\frac{\log^{n+1}m}{n+1}\right).$

One way to evaluate the original sum is to the consider $S_{2^k n}$, a partial sum, and repeatedly split things up with the goal of obtaining the Stieltjes Constants. Here however, we will explore a different approach.

Recall the Dirichlet Eta Function which is given by

$\displaystyle \eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}=\left(1-2^{1-s}\right)\zeta(s).$

Then the $k^{th}$ derivative evaluated at $s=1$ for $\eta(s)$ is equal to the desired sum, that is

$\displaystyle \eta^{(k)}(1)=(-1)^k\sum_{n=1}^\infty \frac{(-1)^{n-1}\log^k n}{n}.$

Now, with this in mind lets find the expansion of $\eta(s)$ around $s=1$. Since

$\displaystyle 1-2^{1-s}=1-e^{-(s-1)\log2}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\left(\log2\right)^{n}}{n!}(s-1)^{n}$

and

$\displaystyle \zeta(s)=\frac{1}{s-1}+\sum_{n=0}^{\infty}(-1)^{n}\gamma_{n}\frac{(s-1)^{n}}{n!},$

by multiplying the two series we have that

$\displaystyle \left(1-2^{1-s}\right)\zeta(s)=\sum_{n=0}^{\infty}\left(\frac{(-1)^{n}\left(\log2\right)^{n+1}}{(n+1)!}+(-1)^{n-1}\sum_{k=0}^{n-1}\gamma_{k}\frac{\log^{n-k}2}{k!(n-k)!}\right)(s-1)^{n}.$

Consequently, our original sum is the $k^{th}$ coefficient above multiplied by $(-1)^k k!$. Specifically, we have that

$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}\log^{k}n}{n}=\frac{\left(\log2\right)^{k+1}}{k+1}-\sum_{j=0}^{k-1}\gamma_{j}\binom{k}{j}\log^{k-j}2.$

For example, when $k=1$, we have that

$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}\log n}{n}=\frac{\log^2(2)}{2}-\gamma \log 2,$

and similarly, when $k=2$,

$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}\log^2 n}{n}=\frac{\log^3 2}{3}-2\gamma \log 2-\gamma_1 \log^2 2.$

I am a PhD candidate in mathematics at Princeton University.
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3 Responses to Evaluating an infinite series in terms of the Stieltjes Constants

1. Chandrasekhar says:

Hey it’s not Steilgies, its Steiljes.

• Eric Naslund says:

Hey thanks Chandru! I see you have a pretty nice blog going there! I am just getting started up (but I have been posting enough)

• Chandrasekhar says:

Hi-
Thanks for your nice opinion. I would like to inform you that there is this link called http://math.is/ which supports MathJaX as a plugin and is very similar to wordpress. You may like to use that site.

Chandrasekhar