# How to evaluate $\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}$ directly?

The rational zeta series $$\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}=\frac{3}{2}-\ln \pi \tag1$$ can be derived from other well known rational zeta series. $$\sum_{n=2}^{\infty}\frac{\left ( -1 \right )^{n}\left ( \zeta (n)-1 \right )}{n+1}=\frac{3}{2}+\frac{\gamma }{2}-\frac{\ln 8\pi}{2} \tag2$$ $$\sum_{n=2}^{\infty}\frac{\zeta (n)-1}{n+1}=\frac{3}{2}-\frac{\gamma }{2}-\frac{\ln 2\pi}{2} \tag3$$ Zeta series (2) and (3) can be derived by integrating the Taylor series of logarithm of gamma function. Zeta series (2)+(3) gives $$\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{2n+1}=\frac{3}{2}-\frac{\ln 4\pi}{2} \tag4$$ The zeta series below can be derived directly with the integral definition of $$\zeta(2n)$$. $$\sum_{n=1}^{\infty}\frac{\zeta (2n)}{(n+1)(2n+1)}=\frac{1}{2} \tag5$$ From zeta series (5) we get $$\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{(2n+1)(2n+2)}=\frac{3}{4}-\ln 2 \tag6$$ Zeta series (6) can be rewritten as $$\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{2n+1}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}=\frac{3}{4}-\ln 2 \tag7$$ Tegether with zeta series (4), we get the result of zeta series (1).

Other than using known results of rational zeta series, how to evaluate zeta series (1) directly with elementary sum of series and integral?

I have tried several ways without success. One of my attempts: $$\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}x^{n+1}=\sum_{n=1}^{\infty}\sum_{k=2}^{\infty}\frac{1}{k^{2n}}\int_{0}^{x}t^{n}dt=\sum_{k=2}^{\infty}\int_{0}^{x}\sum_{n=1}^{\infty}\left ( \frac{t}{k^{2}} \right )^{n}dt \\ =\sum_{k=2}^{\infty}\int_{0}^{x}\frac{t}{k^{2}-t}dt=\sum_{k=2}^{\infty}\left ( k^{2}\ln\frac{k^{2}}{k^{2}-x}-x\right )$$ It seems this attempt won't give any useful result for a closed form, although the sum of this series does converge to $$(3/2-\ln\pi)$$ slowly when setting $$x=1$$.

• Have you tried using $$\log\Gamma(1+x) = -\gamma x + \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k}x^k, \tag*{|x|<1}$$ ? Commented Jul 11, 2020 at 15:58
• Series (8) is different from what's asked in the title. Can we split it into a different question? Commented Jul 11, 2020 at 16:00
• Yes, I have tried to integrate the Taylor series of logarithm of gamma function, but not working. I also tried the integral definition of $\zeta(2n)$, and still without any result. Commented Jul 11, 2020 at 16:06

We write

$$S := \sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{n+1}$$

for the sum to be computed.

1st Solution. We have

\begin{align*} S = \sum_{n=1}^{\infty} \frac{1}{n+1} \sum_{k=2}^{\infty} \frac{1}{k^{2n}} = \sum_{k=2}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n+1} \frac{1}{k^{2n}} = \sum_{k=2}^{\infty} \left( - k^2 \log \left( 1 - \frac{1}{k^2} \right) - 1 \right). \end{align*}

In order to compute this, we write $$S_K$$ for the partial sums of the last step. Then

\begin{align*} S_K &= -K + 1 + \sum_{k=2}^{K} k^2 \log \left( \frac{k^2}{(k+1)(k-1)} \right) \\ &= -K + 1 + \sum_{k=2}^{K} 2 k^2 \log k - \sum_{k=3}^{K+1} (k-1)^2 \log k - \sum_{k=1}^{K-1} (k+1)^2 \log k \\ &= -K + 1 + \log 2 - K^2 \log(K+1) + (K+1)^2 \log K \\ &\quad + \sum_{k=2}^{K} (2 k^2 - (k-1)^2 - (k+1)^2 ) \log k \\ &= -K + 1 + \log 2 - K^2 \log\left(1 + K^{-1}\right) + (2K+1)\log K - 2 \log (K!). \end{align*}

Now by the Stirling's approximation and the Taylor series of $$\log(1+x)$$,

$$2\log (K!) = \left(2K + 1\right) \log K - 2 K + \log(2\pi) + \mathcal{O}(K^{-1})$$

and

$$K^2 \log\left(1 + K^{-1}\right) = K - \frac{1}{2} + \mathcal{O}(K^{-1})$$

as $$K \to \infty$$. Plugging this back to $$S_K$$, we get

$$S_K = \frac{3}{2} - \log \pi + \mathcal{O}(K^{-1})$$

and the desired identity follows by letting $$K\to\infty$$.

2nd Solution. We begin by noting that the Taylor expansion of the digamma function

\begin{align*} \psi(1+z) &= -\gamma + \sum_{k=1}^{\infty} (-1)^{k-1} \zeta(k+1) z^{k} \\ &= -\gamma + \zeta(2) z - \zeta(3) z^2 + \zeta(4) z^3 - \dots, \end{align*}

holds for $$|z| < 1$$. Then by the Abel's Theorem,

\begin{align*} S &= \int_{0}^{1} \sum_{n=1}^{\infty} 2 (\zeta(2n)-1) x^{2n+1} \, \mathrm{d}x \\ &= \int_{0}^{1} x^2 \left( \psi(1+x) - \psi(1-x) - \frac{2x}{1-x^2} \right) \, \mathrm{d}x \\ &= \int_{0}^{1} x^2 \left( \psi(1+x) - \psi(2-x) + \frac{1}{1+x} \right) \, \mathrm{d}x, \tag{1} \end{align*}

where the identity

$$\psi(1+z) = \psi(z) + \frac{1}{z} \tag{2}$$

is used in the last step. Then by using the substitution $$x\mapsto 1-x$$, we get

$$\int_{0}^{1} x^2 \psi(2-x) \, \mathrm{d}x = \int_{0}^{1} (1-x)^2 \psi(1+x) \, \mathrm{d}x.$$

Plugging this back to $$\text{(1)}$$ and performing integration by parts,

\begin{align*} S &= \int_{0}^{1} (2x-1) \psi(1+x) \, \mathrm{d}x + \int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x \\ &= -2 \int_{0}^{1} \log\Gamma(1+x) \, \mathrm{d}x + \int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x. \end{align*}

Now the integrals in the last step can be computed as

$$\int_{0}^{1} \log\Gamma(1+x) \, \mathrm{d}x = -1 + \frac{1}{2}\log(2\pi) \qquad \text{and} \qquad \int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x = -\frac{1}{2} + \log 2.$$

For instance, the first integral can be computed by writing $$\log\Gamma(x+1) = \log\Gamma(x) + \log x$$ and applying the Euler's reflection formula. For a more detail, check this posting. Finally, plugging these back to $$S$$ proves the desired identity.

• Thank you for your detailed derivation for this rational zeta series. My attempt is correct, just lack of skills to continue the evaluation. Commented Jul 12, 2020 at 9:33

Inspired by the above answer of Sangchul, a general solution is derived for this rational zeta series.

Beggining with Digamma function \begin{align*} &\psi(1+z)= -\gamma + \sum_{n=1}^{\infty} (-1)^{n-1} \zeta(n+1) z^{n} \\ &\psi(1-z)= -\gamma - \sum_{n=1}^{\infty} \zeta(n+1) z^{n} \\ & z^2 \left( \psi(1+z) - \psi(1-z) \right)=\sum_{n=1}^{\infty}2\zeta(2n)z^{2n+1} \end{align*} Set $$f(x)$$ as \begin{align*} f(x)&=\sum_{n=1}^{\infty}\frac{\zeta(2n)-1}{n+1}x^{2n} \\ \\ x^2f(x)&=\sum_{n=1}^{\infty}\frac{\zeta(2n)-1}{n+1}x^{2n+2}=\sum_{n=1}^{\infty}2\left( \zeta(2n)-1\right)\frac{x^{2n+2}}{2n+2} \\ &=\sum_{n=1}^{\infty}\left(2\left( \zeta(2n)-1\right)\int_{0}^{x}t^{2n+1}dt \right) \\ &=\int_{0}^{x}\sum_{n=1}^{\infty}\left(2\left( \zeta(2n)t^{2n+1}-t^{2n+1}\right) \right)dt \\ &=\int_{0}^{x} \left( t^2 \left(\psi(1+t) - \psi(1-t) \right) - \frac{2t^3}{1-t^2}\right)dt \\ \end{align*} Performing integration by parts \begin{align*} &\int_{0}^{x}t^2 \psi(1+t)\,dt=x^2\ln\Gamma(1+x)-2\int_{0}^{x}t\ln\Gamma(1+t)\,dt \\ &\int_{0}^{x}t^2 \psi(1-t)\,dt=-x^2\ln\Gamma(1-x)+2\int_{0}^{x}t\ln\Gamma(1-t)\,dt \\ &\int_{0}^{x}\frac{2t^3}{1-t}\,dt=-x^2-\ln(1-x^2) \\ &\Gamma(1+x)\Gamma(1-x)=\frac{\pi x}{\sin(\pi x)} \end{align*} Then \begin{align*} x^2 f(x)&=x^2\ln\frac{\pi x}{\sin(\pi x)}-2\int_{0}^{x}t\ln\frac{\pi t}{\sin(\pi t)}dt+x^2+\ln(1-x^2) \\ &=x^2\ln\frac{\pi x}{\sin(\pi x)}-2\int_{0}^{x}t\ln\pi t\,dt+2\int_{0}^{x}t\ln\sin(\pi t)\,dt+x^2+\ln(1-x^2) \\ &=x^2\ln\frac{\pi x}{\sin(\pi x)}-\frac{x^2}{2}\left(2\ln\pi x -1 \right)+2\int_{0}^{x}t\ln\sin(\pi t)\,dt+x^2+\ln(1-x^2) \\ &=\frac{3}{2}x^2+\ln(1-x^2)-x^2\ln\sin(\pi x)+2\int_{0}^{x}t\ln\sin(\pi t)\,dt \\ &=\frac{3}{2}x^2+\ln(1-x^2)-x^2\ln\sin(\pi x)+2\left( \frac{1}{2}x^2\ln\sin(\pi x)-\frac{\pi}{2}\int_{0}^{x}t^2\cot(\pi t)\,dt \right) \\ &=\frac{3}{2}x^2+\ln(1-x^2)-\pi\int_{0}^{x}t^2\cot(\pi t)\,dt \end{align*} Finally we get the general solution for this rational zeta series as a function of $$x$$. \begin{align*} f(x)&=\sum_{n=1}^{\infty}\frac{\zeta(2n)-1}{n+1}x^{2n} \\ &=\frac{3}{2}+\frac{1}{x^2}\ln(1-x^2)-\ln\sin(\pi x)+\frac{2}{x^2}\int_{0}^{x}t\ln\sin(\pi t)\,dt \\ &=\frac{3}{2}+\frac{1}{x^2}\ln(1-x^2)-\frac{\pi}{x^2}\int_{0}^{x}t^2\cot(\pi t)\,dt \\ \\ f(1)&=\sum_{n=1}^{\infty}\frac{\zeta(2n)-1}{n+1} \\ &=\frac{3}{2}+\lim_{x\rightarrow 1}\left ( \frac{1}{x^2}\ln(1-x^2) -\ln\sin(\pi x))\right )+2\int_{0}^{1}t\ln\sin(\pi t))\,dt \\ &=\frac{3}{2}+\ln\frac{2}{\pi}+2\left ( -\frac{1}{2}\ln2 \right ) \\ &=\frac{3}{2}-\ln\pi \\ \\ f(\frac{1}{2})&=\sum_{n=1}^{\infty}\frac{\zeta(2n)-1}{(n+1)2^{2n}} \\ &=\frac{3}{2}+4\ln\frac{3}{4}-4\pi\int_{0}^{\frac{1}{2}}t^2\cot(\pi t))\,dt \\ &=\frac{3}{2}+4\ln\frac{3}{4}-4\pi\left( \frac{\ln2}{4\pi}-\frac{7\zeta(3)}{8\pi^3} \right) \\ &=\frac{3}{2}+4\ln\frac{3}{4}+\frac{7\zeta(3)}{2\pi^2}-\ln2 \end{align*} Wolfram Alpha gives the closed form of the integrals for $$x=1$$ and $$x=\frac{1}{2}$$.

Integrating the generating sum

$$g(z) = \sum _{n=1}^{\infty } z^n (\zeta (2 n)-1)=\frac{z}{z-1}-\frac{1}{2} \pi \sqrt{z} \cot \left(\pi \sqrt{z}\right)+\frac{1}{2}\tag{1}$$

gives

$$s :=\sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{n+1}= \int_{0}^{1} g(z)\;dz = \frac{3}{2}-\log (\pi )\tag{2}$$

Derivations

Derivation of $$(1)$$

Let $$g(z) = g_1(z) + g_2(z)$$ where

\begin{align}g_1(z) & =\sum _{n=1}^{\infty } z^n \zeta (2 n)=\sum _{n=1}^{\infty } z^n \left(\sum _{k=1}^{\infty } \frac{1}{k^{2 n}}\right)\\ &=\sum _{k=1}^{\infty } \left(\sum _{n=1}^{\infty } \frac{z^n}{k^{2 n}}\right)=\sum _{k=1}^{\infty } \frac{z}{k^2-z}=\frac{1}{2} \left(1-\pi \sqrt{z} \cot \left(\pi \sqrt{z}\right)\right)\end{align}\tag{3}

$$g_2(z) = -\sum _{n=1}^{\infty } z^n = -\frac{z}{1-z}$$

which gives

$$g(z) = g_1(z) + g_2(z) = \frac{z}{z-1}-\frac{1}{2} \pi \sqrt{z} \cot \left(\pi \sqrt{z}\right)+\frac{1}{2}$$

and proves $$(1)$$.

Now to the integral in $$(2)$$.

Starting with the indefinite integration we have

\begin {align} a=\int \left(\frac{z}{z-1}-\frac{1}{2} \pi \sqrt{z} \cot \left(\pi \sqrt{z}\right)+\frac{1}{2}\right) \, dz\\ =-\frac{i \sqrt{z} \text{Li}_2\left(e^{-2 i \pi \sqrt{z}}\right)}{\pi }-\frac{\text{Li}_3\left(e^{-2 i \pi \sqrt{z}}\right)}{2 \pi ^2}-\frac{1}{3} i \pi z^{3/2}+\frac{3 z}{2}\\ -z \log \left(1-e^{-2 i \pi \sqrt{z}}\right)+\log (z-1)+\frac{i \pi }{24} \end{align}\tag{4}

which can be confirmed to be an antiderivative by differentiating. Here $$\text{Li}_n(x)=\sum _{k=1}^{\infty } \frac{x^k}{k^n}$$ is the polylogarithm.

The limits at the integration borders are

$$a_1:=\lim_{z\to 1} a = -\frac{\zeta (3)}{2 \pi ^2}+\frac{3}{2}+\frac{25 i \pi }{24}-\log (\pi )\tag{5a}$$ $$a_0:=\lim_{z\to 0} a =-\frac{\zeta (3)}{2 \pi ^2}+\frac{25 i \pi }{24}\tag{5b}$$

Hence the integral is $$a_1 - a_0$$ where some terms cancel leaving $$\frac{3}{2}-\log (\pi )$$ which proves $$(2)$$.

Notice that $$\log(\pi)$$ enters through this limit

$$\underset{z\to 1}{\text{lim}}\left(\log (z-1)-z \log \left(1-e^{-2 i \pi \sqrt{z}}\right)\right)=-\log (\pi )+\frac{3 i \pi }{2}\tag{6}$$

Discussion

simplified approach

Integrating the generating sum, derived in $$(3)$$ but with the sum left unevaluated in order to avoid integration of the terms containing $$\cot(.)$$

$$g(z) = \sum _{n=1}^{\infty } z^n (\zeta (2 n)-1)=\sum _{k=2}^{\infty } \frac{z}{k^2-z}\tag{a}$$

gives

\begin{align} & s := \sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{n+1}= \int_{0}^{1} g(z)\;dz \\ & = \sum _{k=2}^{\infty }\int_{0}^{1} \frac{z}{k^2-z}\;dz\\ & = \sum _{k=2}^{\infty } (-1 - k^2 (\log(1 - \frac{1}{k^2})))\\ & =\frac{3}{2}-\log (\pi )\end{align}\tag{b}

The last step in $$(b)$$ was carried out with Mathematica. As simple as it seems, I could not derive it by hand. But the result is justified by the original solution above, involving the integration of the $$\cot$$ term.

Note added: I have seen later that the sum in $$(b)$$ was nicely calculated elementarily in the 1st solution of the answer of Sangchul Lee.

Here $$\pi$$ comes from the Stirling appoximation of $$k!$$ for large $$k$$.

approximative form of g(z)

By plotting $$g(z)$$ we see that it is close to the straigt line $$\frac{3}{4} z$$, with a relative mean quadratic deviation of $$0.00259583$$.

Taking this approximation as a generating function gives for the sum $$s$$ the value $$\frac{3}{8}=0.375$$. The exact value is $$\frac{3}{2}-\log(\pi) = 0.35527$$.