Find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $

Question

How can I find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $? It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...

Let consider sum no to infinity, but to n $$ \sum_{k=1}^{n} (k\cdot \ln \frac{2k+1}{2k-1} - 1) =$$ $$ ln \frac{3}{1}\cdot \left(\frac{5}{3}\right)^2 \cdot...\cdot \left(\frac{2n+1}{2n-1}\right)^n - n = \ln \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1} - n $$ but $$ n = \ln e^n $$ so it will be $$ln\frac{1}{e^n} \cdot \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1}$$

So the limit of it is $-\infty$ Have I done this well or I missed sth?

it doesn't converge because the summand is around $n\ln(1+\frac{2}{2n-1})-1 \approx n\frac{2}{2n-1}-1 = \frac{1}{2n-1}$. — mathworker21, Nov 29 '18 at 19:08
It certainly converges, since the $n$'th term is asymptotic to $1/(12 n^2)$. — Robert Israel, Nov 29 '18 at 19:09
Why? In my opinion it should be: $ ln \frac{3}{1}+\ln\left(\frac{5}{3}\right)^2 +...+ \ln\left(\frac{2n+1}{2n-1}\right)^n - n$ — mvxxx, Nov 29 '18 at 19:30
@mathworker21: the $\mathcal{O}(t^2)$ term from $\ln(1+t)$ contributes here. — metamorphy, Nov 29 '18 at 20:15

user · Accepted Answer · 2018-11-29T21:30:39.077

3

We have that

$$\sum_{n=1}^{N} \left(n\cdot \ln \frac{2n+1}{2n-1} - 1\right)=\sum_{n=1}^{N} \left(n\cdot \ln (2n+1)-n\ln (2n-1) - 1\right)=$$

$$=(1\cdot \ln 3-1\cdot \ln1-1)+(2\cdot \ln 5-2\cdot \ln3-1)+(3\cdot \ln 7-3\cdot \ln5-1)+\ldots=$$

$$=-\ln(3\cdot 5\cdot 7\cdot \ldots\cdot (2N-1))+N\cdot\ln(2N+1)-N=$$$$=-\ln\left(\frac{(2N)!}{2^NN!}\right)+N\cdot\ln(2N+1)-N=$$

$$=\ln\left(\frac{(2^N)^2N!N^N}{(2N)!e^N}\right)+N\cdot\ln\left(1+\frac1{2N}\right)$$

and by Stirling's approximation $N!\sim \sqrt{2\pi N}\left(\frac{N}{e}\right)^N$

$$\frac{(2^N)^2N!N^N}{(2N!)e^N}\sim\frac{(2^N)^2N^N}{e^N}\frac{\sqrt{2\pi N}}{\sqrt{4\pi N}}\frac{N^Ne^{2N}}{e^N4^NN^{2N}}=\frac{1}{\sqrt 2}$$

edited Nov 29 '18 at 21:30

answered Nov 29 '18 at 19:21

user

154,566

I edited my post, can you look at this? – mvxxx Nov 29 '18 at 19:25
As noticed the limit exists since the term is asymthotic to $1/(12n^2)$. Starting by the first step indicated you see that the sum telescopes. – user Nov 29 '18 at 19:28
@gimusi: Just was curious how did you precisely say the term is asymptotic to $1/(12n^2)$? – Yadati Kiran Nov 29 '18 at 19:36
@YadatiKiran It is not difficult to see by $\ln \frac{2n+1}{2n-1}=\ln (1+1/2n)-\ln (1-1/2n)$ and then using Taylor's series for log. – user Nov 29 '18 at 19:46
@gimusi: Got it! Thanks. – Yadati Kiran Nov 29 '18 at 19:55
So my result (from the edited post) is bad? – mvxxx Nov 29 '18 at 20:03
As noticed it can’t be $-\infty$ because it converges to a positive value. – user Nov 29 '18 at 20:05
where I made mistake? – mvxxx Nov 29 '18 at 20:06
You get a wrong result because in the telescoping the last (2n+1)^n term doesn’t cancel out. – user Nov 29 '18 at 20:09
By that correction you should obtain an expression similar to mine. Then use Stirling. – user Nov 29 '18 at 20:11
I am trying but how I can use there stirling? I had not use that before – mvxxx Nov 29 '18 at 20:56
I add some detail on that. – user Nov 29 '18 at 20:57
Where are you going to add these details? – mvxxx Nov 29 '18 at 21:16
@mvxxx I've added some step more. Can you conclude form here? – user Nov 29 '18 at 21:31
You have helped me a lot! I have understand all. Thanks again! – mvxxx Nov 29 '18 at 21:46
@mvxxx That's great and the most important thing form me! Thanks :) – user Nov 29 '18 at 21:47

score 2 · Answer 2 · answered Nov 29 '18 at 19:42

2

If I'm not mistaken, the actual sum is $$ \frac{1 - \ln(2)}{2}$$

answered Nov 29 '18 at 19:42

Robert Israel

448,999

Yes that's agree with my result also. – user Nov 29 '18 at 19:56

score 2 · Answer 3 · answered Nov 29 '18 at 19:53

Let $f(x)=\displaystyle\sum_{n=1}^{\infty}\left(n\ln\frac{n+x}{n-x}-2x\right)$ for $x\in(-1,1)$. Then $$f'(x)=\displaystyle\sum_{n=1}^{\infty}\frac{2x^2}{n^2-x^2}=1-\pi x\cot\pi x$$ (termwise differentiation is admissible because of uniform convergence of the latter series in $[-a,a]$ for any $0<a<1$; the second equality is known). Thus, $$f(x)=x-\frac{1}{\pi}\int_{0}^{\pi x}t\cot t\,dt=x(1-\ln\sin\pi x)+\frac{1}{\pi}\int_{0}^{\pi x}\ln\sin t\,dt.$$ Your sum is $f(1/2)=(1-\ln2)/2$.

Felix Marin · Answer 4 · 2018-11-29T21:30:37.557

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{N \in \mathbb{N}_{\geq 1}}$:

\begin{align} &\bbox[#ffd,10px]{\sum_{n = 1}^{N}\bracks{n\ln\pars{2n + 1 \over 2n - 1} - 1}} \\[5mm] = &\ \sum_{n = 1}^{N}n\ln\pars{2n + 1} - \sum_{n = 1}^{N}n\ln\pars{2n - 1} - N \\[5mm] = &\ -N + \sum_{n = 0}^{N}n\ln\pars{2n + 1} - \sum_{n = 0}^{N - 1}\pars{n + 1}\ln\pars{2n + 1} \\[5mm] = &\ -N + N\ln\pars{2N + 1} -N\ln\pars{2} - \sum_{n = 0}^{N - 1}\ln\pars{n + {1 \over 2}} \\[5mm] = &\ -N + N\ln\pars{N + {1 \over 2}} - \ln\pars{\prod_{n = 0}^{N - 1}\bracks{n + {1 \over 2}}} \\[5mm] = &\ -N + N\ln\pars{N + {1 \over 2}} - \ln\pars{\bracks{N - 1/2}! \over \Gamma\pars{1/2}} \\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}&\ -N + N\ln\pars{N + {1 \over 2}} - \ln\pars{\root{2\pi}\bracks{N - 1/2}^{N}\expo{-N + 1/2} \over \root{\pi}} \\[5mm] = &\ -N + N\ln\pars{N + {1 \over 2}} - \ln\pars{2^{1/2}N^{N}\bracks{1 - {1/2 \over N}}^{N} \expo{-N + 1/2}} \\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim} &\ -N + N\ln\pars{N + {1 \over 2}} - \bracks{{1 \over 2}\,\ln\pars{2} + N\ln\pars{N} - N} \\[5mm] = &\ \underbrace{N\ln\pars{1 + {1 \over 2N}}} _{\ds{\stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\ {1 \over 2}}}\ -\ {1 \over 2}\,\ln\pars{2}\label{1}\tag{1} \\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\to} &\ \bbx{1 - \ln\pars{2} \over 2} \approx 0.1534 \end{align}

@mvxxx I add one more line ( line (1) ) where you can see that the first $\displaystyle -N$ cancels with the $\displaystyle -\left(-N\right)$ at the far right and $\displaystyle N\ln\left(N + {1 \over 2}\right) - N\ln\left(N\right) = N\ln\left(1 + {1 \over 2N}\right) \to {\large{1 \over 2}}$ — Felix Marin, Nov 29 '18 at 21:34

Find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $

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