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I ran across a cool series I have been trying to chip away at.

$$\sum_{k=1}^{\infty}\frac{\zeta(2k+1)-1}{k+2}=\frac{-\gamma}{2}-6\ln(A)+\ln(2)+\frac{7}{6}\approx 0.0786\ldots$$

where A = the Glaisher-Kinkelin constant. Numerically, it is approx. $1.282427\ldots$

I began by writing zeta as a sum and switching the summation order

$$\sum_{n=2}^\infty \sum_{k=1}^\infty \frac{1}{(k+2)n^{2k+1}}$$

The first sum is the series for $-n^3\ln(1-\frac{1}{n^2})-n-\frac{1}{2n}$

So, we have $-\sum_{n=2}^\infty \left[\ln(1-\frac{1}{n^2})+n+\frac{1}{2n}\right]$

This series numerically checks out, so I am onto something. At first glance the series looks like it should diverge, but it does converge.

Another idea I had was to write out the series of the series:

$$1/3(1/2)^{3}+1/4(1/2)^{5}+1/5(1/2)^{7}+\cdots +1/3(1/3)^{3}+1/4(1/3)^{5}+1/5(1/3)^{7}+\cdots +1/3(1/4)^{3}+1/4(1/4)^{5}+1/5(1/4)^{7}+\cdots$$

and so on.

This can be written as $$1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots +1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots + 1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots $$

where $x=1/2,1/3,1/4,\ldots$

This leads to the series representation for:

$$\frac{-\ln(1-x^2)}{x^3}-\frac{1}{x}-\frac{x}{2}$$

Since $x$ is of the form $1/n$, we end up with the same series as before.

Now, my quandary. How to finish?. Where in the world does that Glaisher-Kinkelin constant come in, and how can that nice closed from be obtained?. Whether from the series I have above or some other means. As usual, it is probably something I should be seeing but don't at the moment.

The GK constant has a closed form of $$e^{\frac{1}{12}-\zeta^{'}(-1)}$$.

Which means an equivalent closed form would be $\frac{-\gamma}{2}+\ln(2)+6\zeta^{'}(-1)+\frac{2}{3}$

Thanks all.

Michael Hardy
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Cody
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2 Answers2

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We have $$ \begin{eqnarray*} &&\sum_{n=2}^\infty\left( -n^3\log(1-\frac{1}{n^2})-n-\frac{1}{2n}\right)\\ &=&\lim_{N\to\infty}\sum_{n=2}^N\left( -n^3\log(1-\frac{1}{n^2})-n-\frac{1}{2n}\right)\\ &=&\lim_{N\to\infty}\left(\sum_{n=2}^N -n^3\log(1-\frac{1}{n^2})-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}+O\left(\frac{1}{N}\right)\right)\right)\\ &=&\lim_{N\to\infty}[\sum_{n=2}^N (2n^3\log(n)-n^3\log(n+1)-n^3\log(n-1))-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)] \end{eqnarray*} $$ In the sum on the last line, we may gather together the coefficients of each logarithm (terms at the boundary of the sum are a little funny), giving $$ \begin{eqnarray*} &&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)+(N^3+3N^2+3N+1)\log(N)-N^3\log(N+1)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\ &=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)-N^3\log\left(1+\frac{1}{N}\right)+(3N^2+3N+1)\log(N)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\ &=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)-N^3\left(\frac{1}{N}-\frac{1}{2N^2}+\frac{1}{3N^3}+O\left(\frac{1}{N^4}\right)\right)+(3N^2+3N+1)\log(N)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\ &=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\left(3N^2+3N+\frac{1}{2}\right)\log(N)+\left(\frac{-3}{2}N^2+\frac{7}{6}-\frac{\gamma}{2}+\log(2)\right)]\\ &=&-6\log\left(\lim_{N\to\infty}\left(\frac{\prod_{n=1}^N n^n}{N^{N^2/2+N/2+1/12}e^{-N^2/4}}\right)\right)+\frac{7}{6}-\frac{\gamma}{2}+\log(2)\\ &=&-6\log(A)+\frac{7}{6}-\frac{\gamma}{2}+\log(2) \end{eqnarray*} $$ Here, I am taking $$ A=\lim_{N\to\infty}\frac{\prod_{n=1}^N n^n}{N^{N^2/2+N/2+1/12}e^{-N^2/4}} $$ as the definition of the Glaisher-Kinkelin constant.

Julian Rosen
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  • Wow, thanks pink elephants. No wonder I did not finish it:). Thanks a lot. – Cody Jul 24 '12 at 15:24
  • Pink elephants, may I ask another question?. I appreciate your fine response. I follow all of it except for how you got from $\lim_{N\to \infty}\left[\sum_{n=2}^{N}(-2n^{3}\log(n)+n^{3}\log(n+1)+n^{3}\log(n-1))\right]$ to $\lim_{N\to \infty}\left[\sum_{n=2}^{N}(-6\log(n))+\log(2)+(N+1)^{3}\log(N)-N^{3}\log(N+1) \right]$. The part where you mention gathering the coefficients of the logs. Thanks. – Cody Jul 24 '12 at 18:32
  • @Cody: Suppose $2\leq j\leq N-1$. I want to see what the coefficient of $\log(j)$ in this sum is. We get a contribution of $-2j^3$ from the first term when $n=j$, a contribution of $(j-1)^3$ from the second term when $n=j-1$, and a contribution of $(j+1)^3$ from the third term when $n=j+1$. In total, then, the coefficient of $\log(j)$ is $-2j^3+(j-1)^3+(j+1)^3=6j$ (hmm...above, I said that the coefficient was $-6j$. Maybe this is a mistake?). The coefficients of $\log(2)$, $\log(N)$, and $\log(N+1)$ are a little different, as there are fewer values of $n$ in the summation which contribute. – Julian Rosen Jul 24 '12 at 19:01
  • @Cody I just noticed that I dropped that minus sign earlier. Fixing it now – Julian Rosen Jul 24 '12 at 19:02
  • Thanks a lot PE fore the fast response. I was playing around with it by writing out terms and I see what you mean. It was kind of rough to see (at least for me) at first. The whole thing is clever and cool. I see where the 6n and log(2) came from. It was the other two terms I did not see right off. Thanks a bunch. – Cody Jul 24 '12 at 19:19
  • Hi Pink Elephants. May I ask one other question?. If you do not want to bother, I understand. I have worked through this but for some reason I am at a loss to see where $(N+1)^{3}\log(N)-N^{3}\log(N+1)$ comes from. I have listed out terms and tried to see how to get to it, but it does not work out for me. I know it is more than likely something easy and I am being thick-headed. I listed out N and N+1 terms, but still I am too dumb to see it. May I ask how you arrived at that?. – Cody Jul 25 '12 at 14:54
  • Above, we computed the coefficient of $\log(j)$ for $3\leq j\leq N-1$ (in my comment, I said $2\leq j$, which was a typo). For $j=N$, our computation is not quite right, because there is no $n=N+1$ term in the original sum, so we need to add back the $(N+1)^3\log(N)$ term that we mistakenly subtracted. Alternatively, we could compute that the coefficient of $\log(N)$ is $2N^3-(N-1)^3=N^3+3N^2-3N+1$, which is what we have ($-6N\log(N)$ in the sum, plus the extra). Next, there is only one term in the sum giving a $\log(N+1)$ (this is the term $n=N$), and from this term we get $-N^3\log(N+1)$. – Julian Rosen Jul 25 '12 at 17:20
  • Thank you. I think my biggest trouble was getting signs and what not wrong during the whole procedure. – Cody Jul 25 '12 at 17:23
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\begin{align*}\sum\limits_{k=1}^{\infty} \frac{\zeta(2k+1)-1}{2k+3} &= \sum\limits_{k=1}^{\infty}\sum\limits_{n=2}^{\infty} \frac{1}{(2k+3)n^{2k+1}}\\&= \sum\limits_{n=2}^{\infty}n^2\left(\sum\limits_{k=4}^{\infty} \frac{1}{kn^k} - \frac{1}{2}\sum\limits_{k=2}^{\infty} \frac{1}{kn^{2k}}\right)\\&= \sum\limits_{n=2}^{\infty}n^2\left(-\log\left(1-\frac{1}{n}\right)-\frac{1}{n}-\frac{1}{2n^2}-\frac{1}{3n^3} + \frac{1}{2n^2}+\frac{1}{2}\log\left(1-\frac{1}{n^2}\right)\right)\\&= \sum\limits_{n=2}^{\infty}n^2\left(\frac{1}{2}\log\left(\frac{n+1}{n-1}\right) - \frac{1}{n}-\frac{1}{3n^3}\right)\end{align*}

Now, consider the partial sum:

\begin{align*}S_N=\sum\limits_{n=2}^{N}n^2\log\left(\frac{n+1}{n-1}\right) &= \sum\limits_{n=2}^{N} ((n+1)^2 - 2(n+1)+1)\log (n+1) \\& \quad -\sum\limits_{n=2}^{N} ((n-1)^2 + 2(n-1)+1)\log (n-1) \\&= (N+1)^2\log(N+1)+N^2\log N - 2\log 2 + \log \left(\frac{N(N+1)}{2}\right)\\& \quad +2N\log N -2\sum\limits_{n=2}^{N+1} n\log n-2\sum\limits_{n=2}^{N} n\log n\end{align*}

Now, from the definition of Glaisher-Kinkelin constant,

$$\sum\limits_{n=2}^{N} n\log n = \left(\frac{N^2}{2}+\frac{N}{2}+\frac{1}{12}\right)\log N - \frac{N^2}{4} + \log A + o(1)$$

A simple calculation leads to,

\begin{align*}\sum\limits_{n=2}^{N} n^2\left(\frac{1}{2}\log\left(\frac{n+1}{n-1}\right) - \frac{1}{n}-\frac{1}{3n^3}\right)&= \frac{1}{3}\log N + \frac{13}{12} - 2\log 2 - 2\log A - \frac{1}{3}H_N+o(1)\\&= \frac{13}{12} - 2\log 2 - 2\log A - \frac{\gamma}{3}+o(1)\end{align*}

i.e., $$\sum\limits_{k=1}^{\infty} \frac{\zeta(2k+1)-1}{2k+3} = \frac{13}{12} - 2\log 2 - 2\log A - \frac{\gamma}{3}$$

Martin.s
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