Evaluating an infinite series in terms of the Stieltjes Constants

Posted on May 25, 2011 by Eric Naslund

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.

About Eric Naslund

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

  1. Chandrasekhar says:
    June 13, 2011 at 1:13 pm

    Hey it’s not Steilgies, its Steiljes.

    Reply
    • Eric Naslund says:
      June 13, 2011 at 6:22 pm

      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)

      Reply
      • Chandrasekhar says:
        June 14, 2011 at 8:35 am

        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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