A Tough Series $\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$

I have done series with $\zeta(2k)$ and $\zeta(k)$, but I have no idea with this one:

$$\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$$

$\gamma$ is the Euler–Mascheroni Constant.

This value was given by Mathematica. Any hint?

Solution 1:

I solved it myself.

First we note that

$$\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1} = \sum_{n=2}^\infty \sum_{k=1}^\infty \frac{1}{(k+1)n^{2k+1}}=\sum_{n=2}^\infty \left( -\frac{1}{n}- n\log \left( 1-\frac{1}{n^2}\right)\right)$$

Then $$\begin{aligned} \sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1} &=\sum_{n=2}^\infty \left( -\frac{1}{n}- n\log \left( 1-\frac{1}{n^2}\right)\right) \\ &= \lim_{N\to \infty}\sum_{n=2}^N \left( -\frac{1}{n}- n\log \left( 1-\frac{1}{n^2}\right)\right)\\ &= \lim_{N\to \infty} \left[ -H_N+1-\sum_{n=2}^N n \log(n^2-1)+2\sum_{n=2}^Nn\log(n)\right]\\ &= \lim_{N\to \infty} \left[ -H_N+1-\sum_{n=2}^N \left(n\log(n+1) +n\log(n-1)-2n\log(n)\right)\right] \\ &= \lim_{N\to \infty} \Bigg[ -H_N+1+\log(2)-\sum_{n=3}^{N+1}(n-1)\log(n)-\sum_{n=3}^{N-1}(n+1)\log(n) \\ &\quad+\sum_{n=3}^N2n\log(n)\Bigg] \\ &= \lim_{N\to \infty}\left[-H_N-N\log(N+1)-(N-1)\log(N)+2N\log(N)+1+\log(2) \right]\\ &= \lim_{N\to \infty}\left(- \left(H_N-\log N \right)+\log(2)+1-N\log \left( 1+\frac{1}{N}\right)\right)\\&= \lim_{N\to \infty}\left( - \left(H_N-\log N \right)+\log(2)+\mathcal{O}(N^{-1})\right) \end{aligned}$$

Since $\displaystyle \gamma=\lim_{N\to \infty}(H_N-\log(N))$, we get

$$\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1} =-\gamma+\log(2)$$

Solution 2:

This is similar in some respects, but different in others, from the answer provided by Shobhit Bhatnagar. I also thought a detailed explanation of the steps might prove useful.

First we show a binocular telescoping series: $$ \begin{align} \sum_{n=2}^m\log\left(\frac{n^2}{n^2-1}\right) &=\sum_{n=2}^m\log\left(\frac{n}{n-1}\right)+\sum_{n=2}^m\log\left(\frac{n}{n+1}\right)\tag{1a}\\ &=\log(m)+\log\left(\frac2{m+1}\right)\tag{1b}\\[3pt] &=\log\left(\frac{2m}{m+1}\right)\tag{1c} \end{align} $$ Explanation:
$\text{(1a)}$: $\frac{n^2}{n^2-1}=\frac{n}{n-1}\frac{n}{n+1}$
$\text{(1b)}$: evaluate $2$ telescoping sums
$\text{(1c)}$: simplify

Now we use $(1)$ to compute the sum: $$ \begin{align} &\sum_{k=1}^\infty\frac{\zeta(2k+1)-1}{k+1}\\ &=\sum_{k=1}^\infty\frac1{k+1}\sum_{n=2}^\infty\frac1{n^{2k+1}}\tag{2a}\\ &=\sum_{n=2}^\infty\sum_{k=1}^\infty\frac1{k+1}\frac1{n^{2k+1}}\tag{2b}\\ &=\sum_{n=2}^\infty\left(n\log\left(\frac{n^2}{n^2-1}\right)-\frac1n\right)\tag{2c}\\ &=\lim_{m\to\infty}\sum_{n=2}^m\left((m-(m-n))\log\left(\frac{n^2}{n^2-1}\right)-\frac1n\right)\tag{2d}\\ &=\lim_{m\to\infty}\left(m\log\left(\frac{2m}{m+1}\right)-\sum_{n=2}^m\sum_{j=n+1}^m\log\left(\frac{n^2}{n^2-1}\right)-(H_m-1)\right)\tag{2e}\\ &=\lim_{m\to\infty}\left(m\log\left(\frac{2m}{m+1}\right)-\sum_{j=3}^m\sum_{n=2}^{j-1}\log\left(\frac{n^2}{n^2-1}\right)-(H_m-1)\right)\tag{2f}\\ &=\lim_{m\to\infty}\left(m\log\left(\frac{2m}{m+1}\right)-\sum_{j=3}^m\log\left(\frac{2(j-1)}{j}\right)-(H_m-1)\right)\tag{2g}\\ &=\lim_{m\to\infty}\left(m\log\left(\frac{2m}{m+1}\right)-(m-2)\log(2)-\log\left(\frac2m\right)-(H_m-1)\right)\tag{2h}\\[6pt] &=\lim_{m\to\infty}\left(m\log\left(\frac{m}{m+1}\right)+1+\log(2)+\log(m)-H_m\right)\tag{2i}\\[9pt] &=\log(2)-\gamma\tag{2j} \end{align} $$ Explanation:
$\text{(2a)}$: expand $\zeta(2n+1)-1=\sum\limits_{k=2}^\infty\frac1{k^{2n+1}}$
$\text{(2b)}$: switch order of summation
$\text{(2c)}$: $n\log\left(\frac1{1-1/n^2}\right)-\frac1n=\sum\limits_{k=1}^\infty\frac1{(k+1)n^{2k+1}}$
$\text{(2d)}$: write infinite sum as a limit and $n$ as $m-(m-n)$
$\text{(2e)}$: apply $(1)$, the definition of the Harmonic Numbers, and $\sum\limits_{j=n+1}^m1=m-n$
$\text{(2f)}$: switch the order of summation
$\text{(2g)}$: apply $(1)$
$\text{(2h)}$: evaluate the telescoping sum
$\text{(2i)}$: cancel and combine terms
$\text{(2j)}$: evaluate the limits