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.

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About Eric Naslund

I am a PhD candidate in mathematics at Princeton University.
This entry was posted in Analytic Number Theory and tagged , , , , . Bookmark the permalink.

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

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