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I am interested in evaluating $$\sum_{n=2}^{\infty}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$

I am given the solution for the question is $\,\ln (\pi)-\frac{3}{2}\,.$

$$\sum_{n=2}^{\infty}\left(n^2\ln\left(\!1\!-\!\frac{1}{n^2}\!\right)+1\right)=4\ln\left(\!\frac{3}{4}\!\right)+1+9\ln\left(\!\frac{8}{9}\!\right)+1+\ldots$$

Any tricks to solve it?

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    $\begingroup$ Actually now I realise it's not so easy. Maybe Wallis product? $\endgroup$ Oct 31, 2022 at 15:03
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    $\begingroup$ One approach that works is to define $$S(x):=\sum_{n=2}^{\infty}\left(n^2\ln\left(1-\frac{x^2}{n^2}\right)+x^2\right)$$ and evaluate $S’(x)$ via the Mittag-Leffler pole expansion series of $\cot(x)$ then integrate with respect to $x$ and use $S(0)=0$ to find your integration constant and take $x\to 1$ to get the result you are after. $\endgroup$
    – KStarGamer
    Oct 31, 2022 at 15:23
  • $\begingroup$ Note that this is in similar flavour to the questions, (1), (2), and (3), for which you can use less elementary techniques involving products whose answer can be expressed in terms of the Barnes $G$ function and derivatives of the Hurwitz $\zeta$ function. $\endgroup$
    – KStarGamer
    Oct 31, 2022 at 15:27

6 Answers 6

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Here's an elementary proof, assuming that we're allowed to use Stirling's formula: $$\ln (n!) = \sum_{k=2}^n \ln(k) = n\ln n -n + \frac 1 2 \ln n + \frac 1 2 \ln 2\pi + \mathcal O(\frac 1 n)$$

Let us study the partial sums, denoting $u_k=k^2 \ln \left(1-\frac 1 {k^2} \right) + 1$: $$\begin{align} \sum_{k=2}^n u_k &=\sum_{k=2}^n \left(k^2 \ln \left(1-\frac 1 {k^2} \right) + 1 \right) \\ &=\sum_{k=2}^n \left(k^2\ln \left(\frac {(k+1)(k-1)} {k^2}\right) + 1 \right) \\ &=\sum_{k=2}^n \left(k^2 \ln(k+1) + k^2\ln(k-1) - 2k^2\ln k + 1 \right) \\ &=-\ln 2 + n^2\ln(n+1) - (n+1)^2\ln n +\sum_{k=2}^n \left((k-1)^2 \ln k + (k+1)^2\ln k - 2k^2\ln k + 1 \right) \\ &=-\ln 2 + n^2\ln(n+1) - (n+1)^2\ln n +\sum_{k=2}^n \left(2\ln k + 1 \right) \\ &=-\ln 2 + \left(n^2\ln n + n^2 \ln(1+\frac 1 n)\right) - \left(n^2\ln n + 2n\ln n + \ln n\right) \\ &\ \ \ \ +\left(2n\ln n - 2n + \ln n + \ln 2\pi + \mathcal O(\frac 1 n)\right) + (n-1) \\ &=-\ln 2 + n^2 \ln(1+\frac 1 n) - n - 1 + \ln 2\pi + \mathcal O(\frac 1 n) \\ &=n^2 \left(\frac 1 n - \frac 1 {2n^2} + O(\frac 1 {n^3})\right) - n - 1 + \ln\pi + \mathcal O(\frac 1 n) \\ &=- \frac 3 2 + \ln\pi + \mathcal O(\frac 1 n) \end{align}$$

Therefore, as desired: $$\sum_{k=2}^{+\infty} u_k = \ln\pi - \frac 3 2$$

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Just a little simplification (without usage of WolframAlpha) to the nice posted solutions. Using the similar approach, denoting $$f(x)=\sum^\infty_{n=2}n^2\ln\left(1-\frac{x}{n^2}\right)+x$$ and taking the derivative with respect to $x$: $$f'(x)=\sum_{n=2}^\infty\frac{x}{x-n^2}=\sum_{n=1}^\infty\frac{x}{x-n^2}-\frac{x}{x-1}$$ Using $\,\,\displaystyle \pi\cot\pi z=\frac{1}{z}+2\sum_{n=1}^\infty\frac{z}{z^2-n^2}$ $$f'(x)=\frac{\pi}{2}\sqrt x\cot(\pi\sqrt x)-\frac{1}{2}-\frac{x}{x-1}=-\frac{3}{2}+\frac{\pi}{2}\sqrt x\cot(\pi\sqrt x)-\frac{1}{x-1}$$ and $$f(1)=\int_0^1f'(x)dx=-\frac{3}{2}+\lim_{\epsilon\to 0}\int_0^{1-\epsilon}\bigg(\frac{\pi}{2}\sqrt x\cot(\pi\sqrt x)-\frac{1}{x-1}\bigg)dx$$ $$=-\frac{3}{2}+\lim_{\epsilon\to 0}\Big(I_1(\epsilon)+I_2(\epsilon)\Big)\tag{1}$$ where $$I_2(\epsilon)=\int_0^{1-\epsilon}\frac{dx}{1-x}=-\ln\epsilon\tag{2}$$ and $$I_1(\epsilon)=\frac{\pi}{2}\int_0^{1-\epsilon}\sqrt x\cot(\pi\sqrt x)\,dx=\frac{1}{\pi^2}\int_0^{\pi\sqrt{1-\epsilon}}\cot(t)t^2dt$$ Integrating by part, $$I_1(\epsilon)=\frac{t^2\ln\sin t}{\pi^2}\,\bigg|_0^{\pi\sqrt{1-\epsilon}}-\frac{2}{\pi^2}\int_0^{\pi\sqrt{1-\epsilon}}\ln(\sin t)t dt$$ The second term converges at $\epsilon =0$, therefore $$I_1(\epsilon)=\ln\sin(\pi\sqrt{1-\epsilon})-\frac{2}{\pi^2}\int_0^{\pi}\ln(\sin t)t dt+O(\epsilon)$$ $$=\ln\frac{\pi \epsilon}{2}-\frac{2}{\pi^2}\int_0^{\pi}\ln(\sin t)t dt+O(\epsilon)\tag{3}$$ where $$\int_0^{\pi}\ln(\sin t)t dt=\int_0^{\frac{\pi}{2}}\ln(\sin t)tdt+\int_0^{\frac{\pi}{2}}\ln(\cos t)\Big(t+\frac{\pi}{2}\Big)dt$$ $$=\int_0^{\frac{\pi}{2}}\ln(\sin t)t dt+\int_0^{\frac{\pi}{2}}\ln(\sin t)\Big(\frac{\pi}{2}-t\Big) dt+\frac{\pi}{2}\int_0^{\frac{\pi}{2}}\ln(\cos t)dt$$ Using $\displaystyle \int_0^{\frac{\pi}{2}}\ln(\cos t)dt=\int_0^{\frac{\pi}{2}}\ln(\sin t)dt=-\frac{\pi}{2}\ln 2$ $$\int_0^{\pi}\ln(\sin t)t dt=\pi\int_0^{\frac{\pi}{2}}\ln(\sin t)dt=-\frac{\pi^2}{2}\ln2$$ Using (3), we have for $I_1(\epsilon)$ $$I_1(\epsilon)=\ln\frac{\pi \epsilon}{2}+\ln2+O(\epsilon)\tag{4}$$ Putting (2) and (4) into (1), $$f(1)=-\frac{3}{2}+\lim_{\epsilon\to 0}\Big(-\ln\epsilon+\ln\frac{\pi \epsilon}{2}+\ln2+O(\epsilon)\Big)=-\frac{3}{2}+\ln\pi$$

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This answer is just an elaboration on KStarGamer's comment. I myself went for the calculation idea.

Suppose the sum is converging. We have then $$S(x):=\sum_{n=2}^{\infty}\left(n^2\ln\left(1-\frac{x^2}{n^2}\right)+x^2\right)$$ $$\implies S'(x)=\sum_{n=2}^{\infty}\left(-\frac{2x}{n^2}n^2\frac{1}{1-\frac{x^2}{n^2}}+2x\right)$$ \begin{align*} S'(x)&=\sum_{n=2}^{\infty}\left(-2x{\frac{n^2}{n^2-x^2}}+2x\right)\\ &=2x\sum_{n=2}^{\infty}\left({\frac{n^2}{x^2-n^2}}+1\right)\\ &=2x\sum_{n=2}^{\infty}\left({\frac{n^2+x^2-n^2}{x^2-n^2}}\right)\\ &=x^2\left[2x \sum_{n=2}^{\infty}{\frac{1}{x^2-n^2}}\right]\\ &=\pi x^2\left[2\pi x \sum_{n=2}^{\infty}{\frac{1}{(\pi x)^2-(\pi n)^2}}\right]\\ &=\pi x^2\left[-\frac{2x}{\pi(x^2-1)}-\frac{1}{\pi x}+\underbrace{\frac{1}{\pi x}+2\pi x \sum_{n=1}^{\infty}{\frac{1}{(\pi x)^2-(\pi n)^2}}}_{\cot(\pi x)}{}\right]\\ \end{align*} where $\cot(\pi x)$ is expressed using Mittag-Leffler's theorem on meromorphic functions (Look here). We finally get that $$S'(x)=-x+\left(\pi x^2\cot(\pi x)-\frac{2x^3}{x^2-1}\right)$$ By integrating, we get : $$S(x)=-\frac{x^2}2+\int_0^x\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)dt$$ For $x=1$, we get $$S(1)=-\frac{1}{2}+\int_0^1\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)dt=-\frac{1}{2}+\ln(\pi)-1=\ln(\pi)-\frac{3}{2}.$$ The integral $\int_0^1\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)dt$ was evaluated exactly via Wolframalpha.

Thanks to both Angelo and KStarGamer for their corrections and comments.

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    $\begingroup$ Hamdiken, actually, $$\cot(\pi x)=\frac1{\pi x}+2\pi x\sum_{n=1}^{\infty}\frac1{(\pi x)^2-(\pi n)^2}.$$ Hence, $$\frac{1}{\pi x}+2\pi x \sum_{n=2}^{\infty}{\frac{1}{(\pi x)^2-(\pi n)^2}}=\cot(\pi x)-\frac{2x}{\pi\left(x^2-1\right)}.$$ Could you correct it, please ? $\endgroup$
    – Angelo
    Oct 31, 2022 at 16:33
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    $\begingroup$ Moreover, $$S(x)=-\frac{x^2}2+\int_0^x\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)dt\;.$$ and $$\int_0^1\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)dt=\ln(\pi)-1\;.$$ Look at this: wolframalpha.com/… $\text{Hence, }\\S(1)=-\dfrac12+\displaystyle\int_0^1\left(\pi t^2\cot(\pi t)-\dfrac{2t^3}{t^2-1}\right)dt=\ln(\pi)-\dfrac32\;.$ $\endgroup$
    – Angelo
    Oct 31, 2022 at 16:54
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    $\begingroup$ To simplify the final integral, take $t\to 1-t$ then we have $$S(1)=-\frac{1}{2}+\int_{0}^{1} \left(\pi t^2 \cot(\pi t)-\frac{2t^3}{t^2-1}\right)\, dt=-\frac{3}{2}-\ln(2)+\int_{0}^{1} \left(\frac{1}{t} -\pi (1-t)^2\cot(\pi t)\right)\, dt$$ $\endgroup$
    – KStarGamer
    Oct 31, 2022 at 17:16
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    $\begingroup$ I have posted an answer that completes yours. $\endgroup$
    – KStarGamer
    Oct 31, 2022 at 18:40
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(I have partially copied @Hamdiken's answer to make it more complete)

Define the sum $$S(x):=\sum_{n=2}^{\infty}\left(n^2\ln\left(1-\frac{x^2}{n^2}\right)+x^2\right)$$ and differentiate with respect to $x$ to obtain \begin{align*} S'(x)=\sum_{n=2}^{\infty}\left(-2x{\frac{n^2}{n^2-x^2}}+2x\right) &=2x\sum_{n=2}^{\infty}\left({\frac{n^2}{x^2-n^2}}+1\right)\\ &=2x\sum_{n=2}^{\infty}\left({\frac{n^2+x^2-n^2}{x^2-n^2}}\right)\\ &=x^2\left[2x \sum_{n=2}^{\infty}{\frac{1}{x^2-n^2}}\right]\\ &=\pi x^2\left[2\pi x \sum_{n=2}^{\infty}{\frac{1}{(\pi x)^2-(\pi n)^2}}\right]\\ &=\pi x^2\left[-\frac{2x}{\pi(x^2-1)}-\frac{1}{\pi x}+\underbrace{\frac{1}{\pi x}+2\pi x \sum_{n=1}^{\infty}{\frac{1}{(\pi x)^2-(\pi n)^2}}}_{\cot(\pi x)}{}\right]\\ \end{align*} by the Mittag-Leffler pole expansion of $\cot(\pi x)$. We finally get that $$S'(x)=-x+\left(\pi x^2\cot(\pi x)-\frac{2x^3}{x^2-1}\right)$$ Upon integrating, we determine $$S(x)=-\frac{x^2}2+\int_0^x\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)\,dt$$ For $x=1$, we get $$S(1)=-\frac{1}{2}+\int_0^1\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)\,dt\stackrel{t\to 1-t}{=}-\frac{3}{2}-\ln(2)+\int_{0}^{1} \left(\frac{1}{t} -\pi (1-t)^2\cot(\pi t)\right)\, dt $$

Now by the Laurent series for $\cot(\pi x)$, namely, $$\cot(\pi x) = \frac{1}{\pi x} - \frac{2}{\pi x} \sum_{k=1}^{\infty}\zeta(2k) x^{2k}$$ we have $$S(1)=-\ln(2)+\sum_{k=1}^{\infty}\frac{\zeta(2k)}{k(k+1)(2k+1)}$$ Now since for $\Re(s)>1$ we have the following Mellin transform $$\zeta(s)\Gamma(s)=\int_{0}^{\infty}\frac{x^{s-1}}{e^x-1}\,dx$$ we get $$S(1)=-\ln(2)+\int_{0}^{\infty}\frac{1}{e^x-1}\sum_{k=1}^{\infty}\frac{x^{2k-1}}{\Gamma(2k)k(k+1)(2k+1)}=-\ln(2)-2\int_{0}^{\infty}\frac{2+x^2-2\cosh(x)}{x^3(e^x-1)}\,dx$$

We shall now evaluate the integral $$I:=-2\int_{0}^{\infty}\frac{2+x^2-2\cosh(x)}{x^3(e^x-1)}\,dx$$

Write $$\frac{1}{e^x-1}=\sum_{n=1}^{\infty}e^{-nx}$$ so $$I=-2\sum_{n=1}^{\infty}\int_{0}^{\infty} \frac{e^{-n x}}{x^3}\left(2+x^2-2\cosh(x)\right)\,dx$$ Now using the convolution property of the Laplace Transform: $$\int_{0}^{\infty} f(x)\cdot g(x)\, dx=\int_{0}^{\infty} (\mathcal{L}f)(s)\cdot (\mathcal{L}^{-1}g)(s)\, ds$$ with $f(x) = e^{-nx}\left(2+x^2-2\cosh(x)\right)$ and $g(x)=\frac{-2}{x^3}$ we have $$I=\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{2 s^2}{(n+s-1) (n+s)^3 (n+s+1)}\, ds=\int_{0}^{\infty}\frac{s^2+2 s+2}{s(s+1)}+s^2 \psi ^{(2)}(s)\, ds\stackrel{\text{IBP}}{=}\ln(2\pi)-\frac{3}{2}$$ where $\psi$ is the polygamma function. So $S(1)=\ln(\pi)-\frac{3}{2}$ as required. $\square$

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  • $\begingroup$ KStarGamer, how have you obtained that $$\sum_{k=1}^{\infty}\frac{x^{2k-1}}{\Gamma(2k)k(k+1)(2k+1)}=-2\left(\frac{2+x^2-2\cosh(x)}{x^3}\right)\;?$$ $\endgroup$
    – Angelo
    Oct 31, 2022 at 19:46
  • $\begingroup$ @Angelo Use the Maclaurin series $$\sum_{k=1}^{\infty} \frac{x^{2k-1}}{(2k-1)!}=\sinh(x)$$ and integrate this three times with respect to $x$, ensuring you account for the integration constant each time by using the value at $x=0$ each time. (You'll get a factor of $\frac{1}{4}$ in the sum). $\endgroup$
    – KStarGamer
    Oct 31, 2022 at 19:53
  • $\begingroup$ KStarGamer, please, could you explain how you applied IBP in order to calculate $$\int_{0}^{\infty}\frac{s^2+2 s+2}{s(s+1)}+s^2 \psi ^{(2)}(s)\, ds\stackrel{\text{IBP}}{=}\ln(2\pi)-\frac{3}{2}\;?$$ $\endgroup$
    – Angelo
    Oct 31, 2022 at 20:19
  • $\begingroup$ @Angelo the fraction term is evaluated with partial fractions, the polygamma term is evaluated with integration by parts by integrating the polygamma term and differentiating the $s^2$ term repeatedly until you obtain the digamma function. In the limit, you just use special known values of digamma. $\endgroup$
    – KStarGamer
    Oct 31, 2022 at 20:34
  • $\begingroup$ But the integral of the fraction term is divergent, so how can you evaluate it apart from the others? $\endgroup$
    – Angelo
    Oct 31, 2022 at 20:46
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Too long for a comment.

What is interesting is that the partial sums $$S_p=\sum_{n=2}^{p}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$ are given in terms of the gamma function and the derivatives of the zeta function with respect to its first argument.

$$S_p=-\log(2e)+p+\log \big(\Gamma (p)\,\Gamma (p+2)\big)+$$ $$\zeta ^{(1,0)}(-2,p)-2 \zeta ^{(1,0)}(-2,p+1)+\zeta ^{(1,0)}(-2,p+2)+2 \zeta ^{(1,0)}(-1,p)-2 \zeta ^{(1,0)}(-1,p+2)$$ Expanded as a series $$S_p=\log (\pi )-\frac{3}{2}+\frac{1}{2 p}-\frac{1}{4 p^2}+O\left(\frac{1}{p^3}\right)$$ The same happens with

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

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  • $\begingroup$ Wow. How do you realise that?! $\endgroup$
    – Mathxx
    Nov 2, 2022 at 10:22
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    $\begingroup$ @Mathxx. I suspected that it could be something like that and Mathemetica gave me the full answer. $\endgroup$ Nov 2, 2022 at 10:38
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$$\begin{align}\sum^\infty_{n=2}n^2\ln\left(1-\frac{1}{n^2}\right)+1&=\sum^\infty_{n=2}n^2\ln\left(1-\frac{1}{n^2}\right)+\ln(e)=\\&=\sum^\infty_{n=2}\ln\left(1-\frac{1}{n^2}\right)^{n^2}+\ln(e)\;.\end{align}$$ $$\begin{align}\sum^\infty_{n=2}\ln\left(1-\frac{1}{n^2}\right)^{n^2}+\ln(e)&=\sum^\infty_{n=2}\ln\left(\left(1-\frac{1}{n^2}\right)^{n^2}e\right)=\\&=\ln\prod_{n=2}^\infty\left(\frac{n^2-1}{n^2}\right)^{n^2}\!\!e\,.\end{align}$$ Prove that the input of the logarithm is equal to $\pi/e^{3/2}$ and you would be good. Use this paper to get the answer.

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    $\begingroup$ Sure. But this is not an answer because you haven’t demonstrated what this product equals! $\endgroup$
    – FShrike
    Oct 31, 2022 at 15:24
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    $\begingroup$ @FShrike edited the answer :) $\endgroup$ Oct 31, 2022 at 15:41
  • $\begingroup$ The quantity in the paper is $\prod_{n=2}^\infty \left(\frac{n^2-1}{n^2}\right)^{n^2} e$ while you have $\prod_{n=2}^\infty \left(\frac{n^2-1}{n^2}\right)^{n^2} e^{n^2}$ $\endgroup$
    – gist076923
    Oct 31, 2022 at 15:44
  • $\begingroup$ @gst076923, actually it is the same quantity. $\endgroup$
    – Angelo
    Oct 31, 2022 at 16:08
  • $\begingroup$ Ah it was a typo on the answer's front a few edits ago $\endgroup$
    – gist076923
    Oct 31, 2022 at 16:33

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