Evaluate series: $\sum_{k=1}^\infty (-1)^k\left[ k\ln\left(\frac{k^4+2k^3+k^2}{k^4+2k^3+3k^2+2k+2}\right)+\ln\frac{k^2+2k+1}{k^2+1} \right]$

Question

The series is $$\sum_{k=1}^\infty (-1)^k\left[ k\ln\left(\frac{k^4+2k^3+k^2}{k^4+2k^3+3k^2+2k+2}\right)+\ln\left(\frac{k^2+2k+1}{k^2+1}\right) \right] $$ You won't believe it: this has a closed form! It's the beautiful $$4\coth^{-1}(e^\pi)$$ and Wolfram agrees.

By the way, I tried to prove it: the arguments of the $\ln$'s all factor, and with lots of simplifications the summand inside the brackets reduces to $$[\ 2k\ln(k)+2(k+1)\ln(k+1)-(k+1)\ln(k^2+1)-k\ln(k^2+2k+2)\ ] $$ With this new representation and some struggle (it's not difficult, just some algebraic manipulations), I managed to equate the original sum to the sum $$\sum_{k=1}^\infty (-1)^k\left[\ -2k \tanh^{-1}\left(\frac{k+1}{k^2+k+1}\right)- (k+1)\ln\left(\frac{k^2+1}{(k+1)^2}\right) \right] $$ but from here, I couldn't proceed any further. At this point, I even think this made it only worse.

In addition, notice that $$4\coth^{-1}(e^\pi)=2\ln\left(\frac{e^\pi+1}{e^\pi-1}\right)=2\ln\left(\coth\left(\frac\pi2\right) \right)$$

which follows just from the definition of inverse $\coth$, and maybe this is a bit easier to work with.

If anyone has any idea on how to attack this monster, or comes up with a full solution, please share, I look forward to read it.

Answer 1

Without loss of generality, we write the partial sum as:

$$S_{2N}=\sum_{k=1}^{2N}(-1)^k[\ 2k\ln(k)+2(k+1)\ln(k+1)-(k+1)\ln(k^2+1)-k\ln(k^2+2k+2)\ ] $$

and $S_{2N+1}$ can be treated similarly. Next, we group $S_{2N}$ into three parts:

$$S_{2N}=A+B+C $$

where

$$\begin{align}A&=\sum_{k=1}^{2N}(-1)^k[\ 2k\ln(k)+2(k+1)\ln(k+1)\ ]=2(2N+1)\ln(2N+1)\\ \\ B&=-\sum_{k=1}^{2N}(-1)^k[\ (k-1)\ln(k^2+1) +k\ln\left((k+1)^2+1\right)\ ]=-2N\ln\left((2N+1)^2+1\right)\\ \\ C&=-2\sum_{k=0}^{2N}(-1)^k\ln(k^2+1)\end{align}$$

For $C$, we use Abel-Plana formula for alternating sums (derivations can be found in my old post)

$$\begin{align}\boxed{\sum_{k=0}^{2N} (-1)^kf(k)=\frac{f(0)+f(2N)}{2}-\Im\int_0^\infty \frac{f(it)}{\sinh(\pi t)}dt\\ ~~~~~~~~~~~~~~~~~~~~~~~+\sum_{s=1}^m\frac{B_{2s}}{(2s)!}(4^s-1)f^{(2s-1)}(x)\bigg|_{x=2N}+\hat{R_3}}\tag{1}\end{align}$$

where $f(k)=\ln(k^2+1)$, we get

$$C=-\ln\left( (2N)^2+1\right)+2\int_1^\infty \frac{\pi}{\sinh(\pi t)}dt+\sum_{s=1}^m\frac{B_{2s}}{(2s)!}(4^s-1)f^{(2s-1)}(x)\bigg|_{x=2N}+\hat{R_3}$$

Now, we put $A, B, C$ together,

$$\begin{align}S_{2N}&=2(2N+1)\ln(2N+1)-2N\ln\left((2N+1)^2+1\right)-\ln\left( (2N)^2+1\right)\\ \\ &+2\int_1^\infty \frac{\pi}{\sinh(\pi t)}dt+\sum_{s=1}^m\frac{B_{2s}}{(2s)!}(4^s-1)f^{(2s-1)}(x)\bigg|_{x=2N}+\hat{R_3}\end{align}\tag{2}$$

Take the limit for eq.(2), and note that

$$\lim_{N\to\infty} 2(2N+1)\ln(2N+1)-2N\ln\left((2N+1)^2+1\right)-\ln\left( (2N)^2+1\right)=0$$

and

$$\lim_{N\to\infty} \left[\sum_{s=1}^m\frac{B_{2s}}{(2s)!}(4^s-1)f^{(2s-1)}(x)\bigg|_{x=2N}+\hat{R_3}\right]=0\tag{3}$$

Therefore,

$$\boxed{~S=\lim_{N\to\infty}S_{2N}=2\int_1^\infty \frac{\pi}{\sinh(\pi t)}dt=4\coth^{-1}(e^\pi)~~}$$

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Remarks:

For eq.(3), $m\ge1$ is some arbitrary interger, i.e, we set $m=1$

$$f'(2N)=\frac{4N}{1+4N^2}\longrightarrow 0$$

For the tail part, $\hat{R_3}=2\hat{R}(n=N)-\hat{R}(n=2N)$, and

$$\hat{R}(n)=-\Im\int_0^\infty f''(n+i\theta y)\frac{y^2}{e^{2\pi y}-1}dy$$

where $\theta\in (0,1)$, and plug in $f(x)=\ln(1+x^2)$

$$\Im f''(N+i\theta y)=\frac{2(\theta y-1)N}{(N^2+(\theta y-1)^2)^2}+\frac{2(\theta y+1)N}{(N^2+(\theta y+1)^2)^2}\le \frac{4\theta y}{N^3}$$

Since the integral $\int_0^\infty \frac{y^3}{e^{2\pi y}-1}dy$ is convergent, hence,

$$\hat{R_3}\propto\hat{R}(N)\propto \frac1{N^3}\longrightarrow 0$$

Answer 2

Your sum can be rewritten as the following infinite product (exponential plus separation in even and odd terms): $$S=\log{P} \\=\log{\left(\prod_{n=1}^{\infty}\left(\frac{2n^2-2n+1}{2n^2+2n+1} \right)^{2n}\left(\frac{2n+1}{2n-1} \right)^{4n}\left(\frac{(2n+1)^2(2n-1)^2}{4n^2+1} \right) \right)}.$$ First we must note that: $$\prod_{n=1}^{\infty}\left(\frac{(2n+1)^2(2n-1)^2}{4n^2+1}\right) =\frac{1}{\sinh^2{(\frac{\pi}{2})}},\tag{1}$$ which is easy to obtain from the well known infinite products of $\sinh{(x)}$ and $\cosh{(x)}$ (check it). In other hand we are going to consider: $$P_{k}(a)=\prod_{n=1}^{k}\left( \frac{1}{e}\left(\frac{n+a}{n-1+a}\right)^{n} \right),$$ with arbitrary $a$. The partial product is equivalent: $$P_{k}(a)=\frac{(k+a)^{k}}{e^{k}a(a+1)\cdot \cdot \cdot (k-1+a)}=\frac{(k+a)^{k+1}\Gamma{(a)}}{e^{k}\Gamma{(k+1+a)}}\sim\frac{e^{a}\Gamma{(a)}}{\sqrt{2\pi}}(k+a)^{1/2-a} ,$$ Where we have used Stirling's formula in the last step. Putting $a=1/2$ in this last formula we get for first: $$\prod_{n=1}^{\infty}\left(\frac{1}{e}\left(\frac{2n+1}{2n-1}\right)^{n}\right)=\sqrt{\frac{e}{2}},\tag{2}$$ Extending this to the quadratic case $a=(1+it)/2$ with $t \in {\rm I\!R}$ then $\lim |P_{k}(a)|^{2}$ exists and since: $$|\Gamma(a)|^2=\frac{\pi}{\cosh{(\pi t/2)}},$$ we have: $$\prod_{n=1}^{\infty}\left(\frac{1}{e^{2}}\left(\frac{4n^2+4n+t^2+1}{4n^2-4n+t^2+1}\right)^n \right)=\frac{e}{2\cosh({\pi t/2})}\tag{3}.$$ Now combining $(2)$ and $(3)$ we have: $$\prod_{n=1}^{\infty}\left(\left(\frac{4n^2-4n+t^2+1}{4n^2+4n+t^2+1} \right)^{2n}\left(\frac{2n+1}{2n-1} \right)^{4n}\right)=\cosh^2(\frac{\pi t}{2})\tag{4}.$$ Thus using $(1)$ and $(4)$ when $t=1$ gives: $$P=\left(\frac{\cosh{(\pi/2)}}{\sinh{(\pi/2)}}\right)^2=\coth^2{(\pi/2)} \implies S=2\log{(\coth{(\pi/2)})}.$$ Let me know if you need more details to obtain $(1)$.

Remarks: As suggested by @Zima.

1. When $a=\frac{1+it}{2}$: $$\prod_{n=1}^{\infty}\left|\left( \frac{1}{e}\left(\frac{n+a}{n-1+a}\right)^{n}\right)\right|^2=\prod_{n=1}^{\infty}\left(\frac{1}{e^{2}}\left(\frac{4n^2+4n+t^2+1}{4n^2-4n+t^2+1}\right)^n \right)=\frac{e}{2\cosh({\pi t/2})}.$$
2. We have proved: $$P_{1}=\prod_{n=1}^{\infty}\left(\frac{1}{e}\left(\frac{2n+1}{2n-1}\right)^{n}\right)=\sqrt{\frac{e}{2}},$$ and $$P_{2}=\prod_{n=1}^{\infty}\left(\frac{1}{e^{2}}\left(\frac{4n^2+4n+t^2+1}{4n^2-4n+t^2+1}\right)^n \right)=\frac{e}{2\cosh({\pi t/2})},$$ which implies $$\frac{P_{1}^4}{P_{2}^2}=\prod_{n=1}^{\infty}\left(\left(\frac{4n^2-4n+t^2+1}{4n^2+4n+t^2+1} \right)^{2n}\left(\frac{2n+1}{2n-1} \right)^{4n}\right)=\cosh^2(\frac{\pi t}{2})$$

Answer 3

I guess I'm very late for writing an answer now, but please allow me to provide a more elementary way to evaluate this sum. First we have, \begin{align*} S=&\sum_{n=1}^{\infty}{\left( -1 \right) ^n\left[ n\ln \left( \frac{n^4+2n^3+n^2}{n^4+2n^3+3n^2+2n+2} \right) +\ln \left( \frac{n^2+2n+1}{n^2+1} \right) \right]} \\ =&\sum_{n=1}^{\infty}{\left( -1 \right) ^n\left[ n\ln \left( \frac{\left( n+1 \right) ^2n^2}{\left( \left( n+1 \right) ^2+1 \right) \left( n^2+1 \right)} \right) +\ln \left( \frac{\left( n+1 \right) ^2}{n^2+1} \right) \right]} \\ =&\sum_{n=1}^{\infty}{\left( -1 \right) ^n\left[ n\ln \left( \frac{n^2}{n^2+1} \right) +\left( n+1 \right) \ln \left( \frac{\left( n+1 \right) ^2}{\left( n+1 \right) ^2+1} \right) +\ln \left( \frac{\left( n+1 \right) ^2+1}{n^2+1} \right) \right]} \\ =&\sum_{n=1}^{\infty}{\left( -1 \right) ^n\left[ n\ln \left( \frac{n^2}{n^2+1} \right) +\left( n+1 \right) \ln \left( \frac{\left( n+1 \right) ^2}{\left( n+1 \right) ^2+1} \right) \right]}+\sum_{n=1}^{\infty}{\left( -1 \right) ^n\ln \left( \frac{\left( n+1 \right) ^2+1}{n^2+1} \right)} \\ =&L_1+L_2 \end{align*} $L_1$ is actually a telescoping sum, and is easy to evaluate \begin{align*} L_1=&\sum_{n=1}^{\infty}{\left( -1 \right) ^n\left[ n\ln \left( \frac{n^2}{n^2+1} \right) +\left( n+1 \right) \ln \left( \frac{\left( n+1 \right) ^2}{\left( n+1 \right) ^2+1} \right) \right]} \\ =&\sum_{n=1}^{\infty}{\left( \left( -1 \right) ^nn\ln \left( \frac{n^2}{n^2+1} \right) -\left( -1 \right) ^{n+1}\left( n+1 \right) \ln \left( \frac{\left( n+1 \right) ^2}{\left( n+1 \right) ^2+1} \right) \right)} \\ =&\lim_{N\rightarrow \infty} \left( \sum_{n=1}^N{\left( -1 \right) ^nn\ln \left( \frac{n^2}{n^2+1} \right)}-\sum_{n=1}^N{\left( -1 \right) ^{n+1}\left( n+1 \right) \ln \left( \frac{\left( n+1 \right) ^2}{\left( n+1 \right) ^2+1} \right)} \right) \\ =&\lim_{N\rightarrow \infty} \left( \sum_{n=1}^N{\left( -1 \right) ^nn\ln \left( \frac{n^2}{n^2+1} \right)}-\sum_{n=2}^{N+1}{\left( -1 \right) ^nn\ln \left( \frac{n^2}{n^2+1} \right)} \right) \\ =&\lim_{N\rightarrow \infty} \left( \ln \left( 2 \right) -\left( -1 \right) ^{N+1}\left( N+1 \right) \ln \left( \frac{\left( N+1 \right) ^2}{\left( N+1 \right) ^2+1} \right) \right) \\ =&\ln \left( 2 \right) \end{align*} $L_2$ is more complicated, but it is not crazily hard. \begin{align*} L_2=&\sum_{n=1}^{\infty}{\left( -1 \right) ^n\ln \left( \frac{\left( n+1 \right) ^2+1}{n^2+1} \right)} \\ =&\sum_{n=0}^{\infty}{\left( -1 \right) ^n\ln \left( \frac{\left( n+1 \right) ^2+1}{n^2+1} \right)}-\ln \left( 2 \right) \\ =&\sum_{n=0}^{\infty}{\left( \ln \left( \frac{\left( 2n+1 \right) ^2+1}{\left( 2n \right) ^2+1} \right) -\ln \left( \frac{\left( 2n+2 \right) ^2+1}{\left( 2n+1 \right) ^2+1} \right) \right)}-\ln \left( 2 \right) \\ =&\sum_{n=0}^{\infty}{\ln \left( \frac{\left( \left( 2n+1 \right) ^2+1 \right) ^2}{\left( 4n^2+1 \right) \left( 4\left( n+1 \right) ^2+1 \right)} \right)}-\ln \left( 2 \right) \\ =&\ln \left( \prod_{n=0}^{\infty}{\frac{\left( \left( 2n+1 \right) ^2+1 \right) ^2}{\left( 4n^2+1 \right) \left( 4\left( n+1 \right) ^2+1 \right)}} \right) -\ln \left( 2 \right) \\ =&\ln \left( \coth ^2\left( \frac{\pi}{2} \right) \right) -\ln \left( 2 \right) \end{align*} I should elabroate on the product. We got these well-known Weierstrass products of sine and cosine $$ \begin{matrix} \frac{\sin \left( x \right)}{x}=\prod_{n=1}^{\infty}{\left( 1-\frac{x^2}{n^2\pi ^2} \right)}& \cos \left( x \right) =\prod_{n=0}^{\infty}{\left( 1-\frac{4x^2}{\left( 2n+1 \right) ^2\pi ^2} \right)}\\ \end{matrix} $$ We can plug in different arguments (complex-valued ones too) into these expression and get $$ \begin{matrix} \frac{2}{\pi}=\prod_{n=1}^{\infty}{\left( \frac{4n^2-1}{4n^2} \right)}& \frac{2}{\pi}\sinh \left( \frac{\pi}{2} \right) =\prod_{n=1}^{\infty}{\left( \frac{4n^2+1}{4n^2} \right)}& \cos \left( \pi \right) =\prod_{n=0}^{\infty}{\left( 1-\frac{4}{\left( 2n+1 \right) ^2} \right)}& \cosh \left( \frac{\pi}{2} \right) =\prod_{n=0}^{\infty}{\left( \frac{\left( 2n+1 \right) ^2+1}{\left( 2n+1 \right) ^2} \right)}\\ \end{matrix} $$ So we have $$ \begin{matrix} \sinh \left( \frac{\pi}{2} \right) =\prod_{n=1}^{\infty}{\left( \frac{4n^2+1}{4n^2-1} \right)},& -\sinh \left( \frac{\pi}{2} \right) =\prod_{n=0}^{\infty}{\left( \frac{4n^2+1}{4n^2-1} \right)}\\ \end{matrix} $$ From that we have \begin{align*} -\frac{\cosh \left( \frac{\pi}{2} \right) \cosh \left( \frac{\pi}{2} \right)}{\sinh \left( \frac{\pi}{2} \right) \sinh \left( \frac{\pi}{2} \right)}=&\prod_{n=0}^{\infty}{\left( \frac{4n^2-1}{4n^2+1} \right) \left( \frac{4\left( n+1 \right) ^2-1}{4\left( n+1 \right) ^2+1} \right) \frac{\left( \left( 2n+1 \right) ^2+1 \right) ^2}{\left( 2n+1 \right) ^4}} \\ =&\prod_{n=0}^{\infty}{\left( \frac{\left( \left( 2n+1 \right) ^2+1 \right) ^2}{\left( 4n^2+1 \right) \left( 4\left( n+1 \right) ^2+1 \right)} \right) \left( \frac{\left( 2n+1 \right) ^2-4}{\left( 2n+1 \right) ^2} \right)} \\ =&\prod_{n=0}^{\infty}{\left( \frac{\left( \left( 2n+1 \right) ^2+1 \right) ^2}{\left( 4n^2+1 \right) \left( 4\left( n+1 \right) ^2+1 \right)} \right)}\prod_{n=0}^{\infty}{\left( 1-\frac{4}{\left( 2n+1 \right) ^2} \right)} \\ =&\prod_{n=0}^{\infty}{\left( \frac{\left( \left( 2n+1 \right) ^2+1 \right) ^2}{\left( 4n^2+1 \right) \left( 4\left( n+1 \right) ^2+1 \right)} \right)}\,\,\cos \left( \pi \right) \end{align*} $$ \Rightarrow \coth ^2\left( \frac{\pi}{2} \right) =\prod_{n=0}^{\infty}{\left( \frac{\left( \left( 2n+1 \right) ^2+1 \right) ^2}{\left( 4n^2+1 \right) \left( 4\left( n+1 \right) ^2+1 \right)} \right)} $$ So, at the end, the sum is equal to $$ S=L_1+L_2=\ln \left( 2 \right) +\ln \left( \coth ^2\left( \frac{\pi}{2} \right) \right) -\ln \left( 2 \right) =2\ln \left( \coth \left( \frac{\pi}{2} \right) \right) $$

Answer 4

This is not an answer.

This only wants to show a proof of Stirling's formula using Euler's infinite product for $\sin{x}$. We have the following: $$\int_{0}^{1}\log{\left( \frac{\sin{(\pi x)}}{\pi x (1-x^2)} \right)}dx=3-\log{(8\pi)}\tag{1}.$$ This integral is elementary. The reader should have not problems to check it. In other hand we have: $$ \frac{\sin{(\pi x)}}{\pi x (1-x^2)}=\prod_{n=2}^{\infty}\left(1-\frac{x^2}{n^2} \right)\tag{2}.$$ The infinite product has uniform convergence from Weierstrass's theorem for infinite products. So combining $(1)$ and $(2)$ and using uniform convergence we can confirm that: $$\int_{0}^{1}\log{\left(\prod_{n=2}^{\infty}\left(1-\frac{x^2}{n^2} \right)\right)}dx=\int_{0}^{1}\sum_{n=2}^{\infty}\log{\left(1-\frac{x^2}{n^2} \right)}dx\\ =\sum_{n=2}^{\infty}\int_{0}^{1}\log{\left(1-\frac{x^2}{n^2} \right)}dx=3-\log{(8\pi)}.$$ Since: $$\int_{0}^{1}\log{\left(1-\frac{x^2}{n^2} \right)}dx=\left((x-n)\log{(\frac{x-n}{n})}+(x+n)\log{(\frac{x+n}{n})} -2x \right) \Big|_{0}^{1},$$ we have: $$\sum_{n=2}^{\infty}\left( (n-1)\log{(\frac{n}{n-1})}+(n+1)\log{(\frac{n+1}{n})} -2 \right)=3-\log{(8\pi)},$$ exponential in this last gives: $$ \prod_{n=2}^{\infty}\left( \frac{1}{e^{2}}\left(1-\frac{1}{n^2}\right)\left(\frac{n+1}{n-1} \right)^{n} \right)=\frac{e^{3}}{8\pi}\tag{3}. $$ Since the following product is telescopic: $$\lim_{N}\prod_{n=2}^{N}\left(1-\frac{1}{n^2}\right)=\lim_{N}\prod_{n=2}^{N}\left(\frac{(n+1)(n-1)}{n\cdot n}\right)=\lim_{N}\frac{N+1}{2N}=\frac{1}{2}.$$ using this last in $(3)$ we have: $$\prod_{n=2}^{\infty}\left( \frac{1}{e^{2}}\left(\frac{n+1}{n-1} \right)^{n} \right)=\frac{e^{3}}{4\pi}.\tag{4}$$ which is equivalent to Stirling's formula.

If we consider the partial product of $(4)$ as $N$ tends to infinity, we have: $$\lim_{N}P_{N}=\lim_{N}\prod_{n=2}^{N}\left( \frac{1}{e^{2}}\left(\frac{n+1}{n-1} \right)^{n} \right)\\=e^{2-2N}2^{-3}3^{-2}4^{-2}\cdot \cdot \cdot (N+1)^{-2}N^{N-1}(N+1)^{N}\\=\lim_{N}\frac{1}{2}\frac{e^{2-2N}N^{N+1}(N+1)^{N}}{N!^2}=\frac{e^{3}}{4\pi},$$ so: $$\lim_{N}\frac{N^{N+1}(N+1)^{N}}{e^{2N}N!^2}=\frac{e}{2\pi},$$ taking the square root in this last: $$\lim_{N}\frac{N^{N/2+1/2}(N+1)^{N/2}}{e^{N}N!}=\frac{\sqrt{e}}{\sqrt{2\pi}}.$$ Finally since $\lim_{N}\left({\frac{1+N}{N}}\right)^{N}=e$, then: $$\lim_{N}\frac{N^{N/2+1/2}(N+1)^{N/2}}{\left({\frac{1+N}{N}}\right)^{N/2}e^{N}N!}=\frac{1}{\sqrt{2\pi}}\implies \lim_{N}\frac{e^{N}N!}{N^{N+1/2}}=\sqrt{2\pi}.$$ And we are done.

An elementary proof for the Euler's product was done by Venkatachaliengar in 1962.

1. Venkatachaliengar. K: "Elementary proofs of the infinite product for $\sin{z}$ and allied formulae". American Mathematical Monthly, 69, 1962, 541-545.