Working with some integrals I stumbled upon the following slowly converging series:

$$ S = \sum_{n = 0}^{\infty}\left(-1\right)^{n} \left[n + \frac{3}{2} + \left(n + 1\right)\left(n + 2\right) \log\left(1 - \frac{1}{n + 2}\right)\right] $$

I have reasons to suspect that the series has a closed form:

$$S=\frac{1}{2}-\frac{7 \zeta(3)}{2 \pi^2}=0.073721601182494209 \ldots$$

The actual proof eludes me so far.

Can you prove or disprove this conjecture?

Writing the logarithm as a series we have:

\begin{align} &\left(n + 2\right) \log\left(1 - \frac{1}{n+2}\right) = -\sum_{k = 1}^{\infty} \frac{1}{k\left(n + 2\right)^{k - 1}} \\ = &\ -1-\sum_{k=1}^\infty \frac{1}{(k+1) (n+2)^k} \end{align} Which turns the series into:

$$S=\sum_{n=0}^\infty (-1)^n \left(\frac{1}{2}-(n+1) \sum_{k=1}^\infty \frac{1}{(k+1) (n+2)^k} \right)$$

I can provide the way I came to this expression, but it's very long and complicated, as usual. I'd like some clear proof, if possible.

To get the sense of how slowly the series converges, for $20000$ terms the result agrees with the stated closed form in $4$ first significant digits.

The integral from which this series was obtained is (again, conjectured):

$$\int_0^1 {_2 F_1} (1,-t;2-t;-1) dt = \frac{7 \zeta(3)}{\pi^2}+\frac{1}{2}$$

I don't think it's very useful, except for numerical confirmation.

  • $\begingroup$ Since you were working with some integrals, do you have an equivalent integral representation for that sum? $\endgroup$
    – Zacky
    Aug 10, 2019 at 19:00
  • 1
    $\begingroup$ By Maple it agrees for at least $200$ digits, although it does not compute the closed form. If there was an integral form though, I think it could.(not saying it would be a proof, but still verification) $\endgroup$
    – Sil
    Aug 10, 2019 at 19:02
  • 1
    $\begingroup$ @ㄴㄱ, sure, please see the edit $\endgroup$
    – Yuriy S
    Aug 10, 2019 at 19:03
  • $\begingroup$ As it is already pointed out in one of the answers, the best way to attack such a problem is by functional equations. From seeing the conjectured result this was my first bet too, since the combination of Apéry's constant divided by $\pi^2$ is a rather strange one apart from derivatives of zeta-like functions via the functional equations (at least from what I have experienced). $\endgroup$
    – mrtaurho
    Aug 11, 2019 at 12:20
  • $\begingroup$ $\texttt{Relative Error}$ is $\displaystyle\large 10^{-13} \%$. $\endgroup$ Aug 11, 2019 at 20:09

2 Answers 2


For $\Re(s)> 0$, let $F(s)$ be defined as$$ \small F(s) = \sum_{n=0}^\infty (-1)^n\left[(n+1)^{1-s} + \frac 1 2 (n+1)^{-s} + \color{red}{(n+1)(n+2)}\left[(n+1)^{-s}\ln(n+1) - \color{red}{(n+2)^{-s}\ln(n+2)}\right]\right]. $$ Then $F(s)$ is analytic and $\displaystyle S= \lim_{\substack{s\to 0\\\Re(s)>0}}F(s)$ holds. Assume $\Re(s)>3$ for a moment so that each term in the summand is absolutely summable. Then by making $n+1\mapsto n$ to the red-colored term, we get \begin{align*} F(s) =& \sum_{n=0}^\infty (-1)^n\left[(n+1)^{1-s} + \frac 1 2 (n+1)^{-s} + 2(n+1)^{2-s} \ln(n+1)\right]\\ =&\eta(s-1) +\frac 1 2 \eta(s) -2\eta'(s-2) ,\qquad \Re(s)>3 \end{align*} where $\eta(s)$ is the Dirichlet's eta function. By analytic continuation, this should also hold for all $\Re(s)>0$ and it follows $$ S = \lim_{\substack{s\to 0\\\Re(s)>0}}F(s) = \eta(-1) + \frac 1 2 \eta(0)-2\eta'(-2)=\frac 1 2 - \frac{7\zeta(3)}{2\pi^2} $$ where the result can be derived from the functional equations satisfied by $\eta(s)$ and $\zeta(s)$, i.e. $$ \eta(s) = (1-2^{1-s})\zeta(s),\qquad \zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s). $$

  • $\begingroup$ Great answer - compact and elegant $\endgroup$
    – Klangen
    Jan 7, 2020 at 7:06

Splitting up the positive and negative terms gives


Since $\displaystyle\ln\left(1-\frac1n\right)=\ln(n-1)-\ln(n)=-\int_0^1\frac1{x+n-1}~\mathrm dx$ we provide the more suitable form

$$S=\int_0^1\sum_{n=1}^\infty-1+\frac{2n(2n+1)}{x+2n}-\frac{2n(2n-1)}{x+2n-1}~\mathrm dx$$

The summand can be managed with some long division and the digamma function, reducing this down to

$$S=\int_0^1x(x-1)\sum_{n=1}^\infty\frac{(-1)^n}{x+n}~\mathrm dx\\=\frac12\int_0^1x(1-x)\left[\psi^{(0)}\left(\frac{x+2}2\right)-\psi^{(0)}\left(\frac{x+1}2\right)\right]~\mathrm dx$$

Substituting $x\mapsto1-x$ in the second digamma function and using the reflection formula gives us

$$S=\frac12\int_0^1x(1-x)\left[\frac2x-\pi\cot\left(\frac{\pi x}2\right)\right]~\mathrm dx$$

which reduces to a few polylogarithms and the rest is plugging and chugging.

  • $\begingroup$ Nice! By the way last integral Maple handles by convert((1/2)*(int(x*(1-x)*(2/x-Pi*cot((1/2)*Pi*x)), x = 0 .. 1)), Elliptic_related) and gives $\frac{1}{2}-\frac{7 \zeta(3)}{2 \pi^2}$ . $\endgroup$
    – Sil
    Aug 10, 2019 at 21:04

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