# How to calculate $\sum_{n= 0}^{\infty} \left( \frac {n+1} {n+2}+n(n+1)\ln \left(1-\frac 1{(n+1)^2}\right)\right)$

How to calulate $$\sum_{n= 0}^{\infty} \left( \frac {n+1} {n+2}+n(n+1)\ln \left(1-\frac 1{(n+1)^2}\right)\right)?$$

My goal was to calculate the integral $$\int_{0}^{\infty}\left[\frac{1}{x}-\frac{1}{e^x-1}\right]^2 dx$$.

Using the three results

1) $$\frac{1}{(1-e^{-x})^2}=\sum_{n=0}^{\infty}(n+1)e^{-nx}$$

2)$$a_n=\int_0^1\int_0^1\frac{1}{(n+u+v)^3}uvdudv=\frac{1}{2(n+2)}+\frac{n}{2}\log\left(1-\frac{1}{(n+1)^2}\right)$$

3)$$e^x-1-x=e^xx^2\int_0^1ue^{-ux}du$$.

We deduce

$$\int_{0}^{\infty}\left[\frac{1}{x}-\frac{1}{e^x-1}\right]^2dx$$

=$$\int_{0}^{\infty}\frac{x^2}{(1-e^{-x})^2}\left[\int_0^1ue^{-ux}du\int_0^1ve^{-vx}dv\right]dx$$

=$$\int_0^1\int_0^1\left[\sum_0^{\infty}(n+1)\int_0^{\infty}x^2e^{-x(n+u+v)}dx\right]uvdudv$$

= $$2\int_0^1\int_0^1\left[\sum_0^{\infty}\frac{n+1}{(n+u+v)^3}\right]uvdudv$$

= $$2\sum_0^{\infty}(n+1) a_n =?$$

I'm interested in other ways too, but I'd like to understand how to calculate the sum of this series

• Do you want to solve the integral in this particular direction? Because there are other ways to solve the integral. – Sangchul Lee Jul 8 '19 at 23:09
• I'm interested in other ways too, but I'd like to understand how to calculate the sum of this series – Tina Jul 8 '19 at 23:18

By writing $$\log\left(1-\frac{1}{(n+1)^2}\right)=\log n+\log(n+2)-2\log(n+1)$$, we get

\begin{align*} &\sum_{n=1}^{N} n(n+1)\log\left(1-\frac{1}{(n+1)^2}\right) \\ &= \sum_{n=1}^{N} n(n+1)\log n - 2 \sum_{n=1}^{N} n(n+1)\log(n+1) + \sum_{n=1}^{N} n(n+1)\log(n+2) \\ &= \Bigg( \sum_{n=1}^{N} n(n+1)\log n - 2 \sum_{n=1}^{N} (n-1)n\log n + \sum_{n=1}^{N} (n-2)(n-1)\log n \Bigg) \\ &\qquad - 2N(N+1)\log(N+1) + N(N+1)\log (N+2) + (N-1)N\log(N+1) \\ &= 2\log(N!) - N(N+3)\log(N+1) + N(N+1)\log(N+2). \end{align*}

Now expanding the last expression for large $$N$$ and using the Stirling's approximation, we easily check that

\begin{align*} \begin{gathered} 2\log(N!) = (2N+1)\log N - 2N + \log(2\pi) + o(1), \\ - N(N+3)\log(N+1) = -N(N+3)\log N - N - \frac{5}{2} + o(1), \\ N(N+1)\log(N+2) = N(N+1)\log N + 2N + o(1), \end{gathered} \end{align*}

as $$N\to\infty$$, and so,

\begin{align*} \sum_{n=1}^{N} n(n+1)\log\left(1-\frac{1}{(n+1)^2}\right) = \log N - N + \log(2\pi) - \frac{5}{2} + o(1) \quad \text{as } N \to \infty. \end{align*}

On the other hand,

$$\sum_{n=0}^{N} \frac{n+1}{n+2} = N+2 - \sum_{n=1}^{N+2} \frac{1}{n} = N+2-(\gamma + \log(N+2) + o(1))$$

by the definition of the Euler-Mascheroni constant $$\gamma$$. Combining altogether and letting $$N\to\infty$$, we get

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

Here, we regard $$0\log 0 = 0$$. The following numerical computation confirms this:

Addendum. We may regularize the integral by

$$I(\alpha, s) := \int_{0}^{\infty} x^{s-1} \left(\frac{1}{x} - \frac{\alpha}{e^{\alpha x} - 1} \right)^2 \, \mathrm{d}x.$$

This integral convergess for $$\alpha > 0$$ and $$s \in (0, 2)$$. Also, the substitution $$\alpha x \mapsto x$$ gives

$$I(\alpha, s) = \alpha^{2-s} I(1, s).$$

Our goal is to find a closed form of $$I(1, s)$$ and utilize it to compute the value of $$I(1, 1)$$. We begin by noting that

$$\int_{0}^{\infty} \frac{x^{s-1}}{e^{\alpha x} - 1} \, \mathrm{d}x = \frac{\Gamma(s)}{\alpha^s}\zeta(s)$$

for any $$\alpha > 0$$ and $$s > 1$$. Differentiating both sides by $$\alpha$$,

$$-\int_{0}^{\infty} \frac{x^{s}e^{\alpha x}}{(e^{\alpha x} - 1)^2} \, \mathrm{d}x = -\frac{\Gamma(s+1)}{\alpha^{s+1}}\zeta(s).$$

Replacing $$s$$ by $$s-1$$ and subtracting $$\int_{0}^{\infty} \frac{x^{s-1}}{e^{\alpha x} - 1} \, \mathrm{d}x$$ from both sides,

$$\int_{0}^{\infty} \frac{x^{s-1}}{(e^{\alpha x} - 1)^2} \, \mathrm{d}x = \frac{\Gamma(s)}{\alpha^{s}}[ \zeta(s-1) - \zeta(s)],$$

which we know to hold for any $$\alpha > 0$$ and $$s > 2$$. On the other hand,

\begin{align*} I(1,s) - I(\alpha, s) &= \int_{0}^{\infty} x^{s-1} \bigg[ \left(\frac{1}{x} - \frac{1}{e^{x} - 1} \right)^2 - \left(\frac{1}{x} - \frac{\alpha}{e^{\alpha x} - 1} \right)^2 \bigg] \, \mathrm{d}x \\ &= \int_{0}^{\infty} x^{s-1} \bigg( -\frac{2x^{-1}}{e^{x} - 1} + \frac{1}{(e^{x} - 1)^2} + \frac{2\alpha x^{-1}}{e^{\alpha x} - 1} - \frac{\alpha^2}{(e^{\alpha x} - 1)^2} \bigg) \, \mathrm{d}x. \end{align*}

We notice that this integral converges for all $$s > 0$$. Moreover, for each $$\alpha > 0$$ this defines a holomorphic function in $$s > 0$$. Now by assuming $$s > 2$$,

\begin{align*} I(1,s) - I(\alpha, s) &= \int_{0}^{\infty} \bigg( -\frac{2x^{s-2}}{e^{x} - 1} + \frac{x^{s-1}}{(e^{x} - 1)^2} + \frac{2\alpha x^{s-2}}{e^{\alpha x} - 1} - \frac{\alpha^2 x^{s-1}}{(e^{\alpha x} - 1)^2} \bigg) \, \mathrm{d}x \\ &= (1 - \alpha^{2-s})\big( \Gamma(s)[ \zeta(s-1) - \zeta(s)] - 2\Gamma(s-1)\zeta(s-1) \big). \end{align*}

Although this holds a priori for $$s > 2$$, this continues to hold for all of $$s > 0$$ by the principle of analytic continuation. Comparing this with $$I(1,s) - I(\alpha, s) = (1 - \alpha^{2-s})I(1, s)$$, we get

$$I(1, s) = \Gamma(s)[ \zeta(s-1) - \zeta(s)] - 2\Gamma(s-1)\zeta(s-1)$$

for $$s \in (0, 2)$$. Using $$\zeta(s) = \frac{1}{s-1} + \gamma + \mathcal{O}(s-1)$$ as $$s \to 1$$ and $$\zeta(0) = -\frac{1}{2}$$, the right-hand side expands as

$$I(1, s) = \Gamma(s) \bigg( \zeta(s-1) - \gamma - 2 \frac{\zeta(s-1) - \zeta(0)}{s-1} \bigg) + \mathcal{O}(s-1).$$

Letting $$s \to 1$$, this converges to

$$I(1, 1) = \zeta(0) - \gamma - 2\zeta'(0) = -\frac{1}{2} - \gamma + \log(2\pi).$$

• Thank's . Very interesting – Tina Jul 8 '19 at 23:52
• If some one have other way to find the value of integral. I'm interesting also – Tina Jul 8 '19 at 23:55
• Very elegant solution, for sure ! By the way, congratulations for the 100k. Cheers :-) – Claude Leibovici Jul 9 '19 at 6:10
• @Sangchul Lee thank you also for this beautiful second method – Tina Jul 9 '19 at 6:46
• @ClaudeLeibovici, Thank you :) – Sangchul Lee Jul 9 '19 at 12:58

I will use a generalisation of the gamma function, called $$Q_m(x)$$ in the script here . This script is not perfect, but the formulas I use are correct.

We have on page $$13$$ the formula $$(4.1)$$

$$\sum\limits_{n=1}^N n^m \ln\left(1+\frac{x}{n}\right) = p_{m,N}(x) + r_m(x)\ln N - \ln Q_{m,N}(x)$$

$$\displaystyle Q_m(x):=Q_{m,N}(x)|_{N\to\infty} \enspace, \enspace\enspace Q_m^*(x) := (1+x) Q_m(x) \enspace,$$

$$\displaystyle r_1(x)=-\frac{x^2}{2} \enspace, \enspace\enspace r_2(x)=\frac{x^3}{3} \enspace,$$

$$\displaystyle p_{1,N}(x)=xN \enspace, \enspace\enspace p_{2,N}(x)=x\frac{N(N+1)}{2}-\frac{x^2}{2}N \enspace$$ .

Now we can write

$$\displaystyle \sum\limits_{n=0}^{N-1}\left(x^2\frac{n+1}{n+2}+n(n+1)\ln\left(1-\frac{x^2}{(n+1)^2}\right)\right) = x^2(N+1-H_{N+1}) -$$

$$\enspace\enspace - (p_{1,N}(-x) + p_{1,N}(x)) + (p_{2,N}(-x) + p_{2,N}(x)) - (r_1(-x) + r_1(x))\ln N$$

$$\enspace\enspace + (r_2(-x) + r_2(x))\ln N + \ln(Q_{1,N}(-x) Q_{1,N}(x)) - \ln(Q_{2,N}(-x) Q_{2,N}(x))$$

$$\displaystyle \to ~ x^2(1-\gamma) + \ln(Q_1(-x)Q_1(x)) - \ln(Q_2(-x)Q_2(x)) \enspace$$ for $$\enspace N\to\infty$$

$$\displaystyle = x^2(1-\gamma) + \ln(Q_1^*(-x) Q_1(x)) - \ln(Q_2^*(-x) Q_2(x))$$

Formula $$(4.7)$$ on page $$16$$ with $$m:=2$$ leads to $$\ln(Q_2^*(-1)Q_2(1))=\frac{1}{2}-\ln(2\pi)$$ and formula $$(4.16)$$ on page $$24$$ with $$m:=1$$ gives us $$\ln(Q_1^*(-1)Q_1(1))=-1$$

so that with $$x:=1$$ we get finally:

$$\sum\limits_{n=0}^\infty\left(\frac{n+1}{n+2}+n(n+1)\ln\left(1-\frac{1}{(n+1)^2}\right)\right) = -\gamma - \frac{1}{2} + \ln(2\pi)$$