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As title says, does the following series converge? $$\sum_{1}^{\infty}\big[1 + n\ln\big(\frac{2n-1}{2n+1}\big) \big]$$

It is supposed to converge but I don't know how.

My attempt:

$$\sum_{1}^{\infty}\big[1 + n\ln\big(\frac{2n-1}{2n+1}\big) \big] = \sum_{1}^{\infty}a_n,$$ where

\begin{align} a_n &= \big[1 + n\ln\big(\frac{2n-1}{2n+1}\big) \big] \\ &= 1 + n\big[\ln(2n-1) - \ln(2n+1)\big]. \end{align}

I couldn't come up with any good convergence tests and started listing out a few terms:

$a_1 = 1 + 1\ln(1) - 1\ln(3)$,

$a_2 = 1 + 2\ln(3) - 2\ln(5)$,

$a_3 = 1 + 3\ln(5) -3\ln(7)$,

and so on.

Since some terms cancel, I think for the $N$-th partial sum, we have:

\begin{align} \sum_{1}^{N}a_n &= N + \ln(2N-1) - N\ln(2N+1) \\ &= N + \ln\frac{2N-1}{(2N+1)^N} \end{align} At this point, I can only see divergence as $N \rightarrow \infty$.

What am I doing wrong? How can I correctly test for convergence in this case?

js9
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    I would apply the integral test on $\sum_{n=1}^{\infty}\bigg[n\ln\Big(\frac{2n+1}{2n-1}\Big)-1\bigg]$. – Matthew H. Feb 02 '21 at 06:36
  • Wouldn’t the constant term 1 be a problem when applying the integral test? – js9 Feb 02 '21 at 06:38
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    No. The function $f(x)=x\ln\Big(\frac{2x+1}{2x-1}\Big)-1$ is positive and decreasing on $[1,\infty)$ – Matthew H. Feb 02 '21 at 06:40
  • FYI: $$\int \Big[x\ln\Big(\frac{2x+1}{2x-1}\Big)-1\Big]dx=\frac{1}{2}x\Big[x\ln\Big(\frac{2x+1}{2x-1}\Big)-1\Big]+\frac{1}{8}\ln\Big(\frac{2x-1}{2x+1}\Big)+C$$ – Matthew H. Feb 02 '21 at 06:46
  • @MatthewPilling why is there a sign change on 1? Is this important? – js9 Feb 02 '21 at 06:47
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    I multiplied what you called $a_n$ in your post by $-1$ in order to get my summand. This ensures that we have a positive decreasing sequence which is a requirement to use integral test. Once we show that $\sum-a_n$ converges then the convergence of $\sum a_n$ is immediate. – Matthew H. Feb 02 '21 at 06:50
  • @MatthewPilling Oh, sorry I didn’t see the fraction was also reversed. I will try this, thank you! – js9 Feb 02 '21 at 06:55
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    FYI If you want to evaluate $\lim_{x \rightarrow \infty}x\Big[x\ln\Big(\frac{2x+1}{2x-1}\Big)-1\Big]$ I suggest you perform the substitution $u=\frac{2x+1}{2x-1}$ and note $u\rightarrow 1^+$ as $x \rightarrow \infty$. – Matthew H. Feb 02 '21 at 07:01
  • @js9 You may obtain the upper/lower bounds for $a_n$. – River Li Feb 02 '21 at 07:14
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    There's even an exact evaluation of this sum. (And the question about convergence is asked already, too.) – metamorphy Feb 02 '21 at 07:44
  • @metamorphy, yes actually they both do. Sorry I couldn't find them. – js9 Feb 02 '21 at 19:06

1 Answers1

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Let

$$f(x)=\frac{1}{x}\left(\log\frac{1-\frac x2}{1+\frac x2}\right)+1$$

Then

$$f(x)=\frac1x\log(1-\frac x2)-\frac1x\log(1+\frac x2)+1$$

Use Taylor's formula:

$$\log(1-x)=-x-\frac12x^2-\frac13x^3+O(x^4)$$

$$f(x)=(-\frac12-\frac18x-\frac1{24}x^2+O(x^3))-(\frac12-\frac18x+\frac1{24}x^2+O(x^3))+1$$

$$f(x)=-\frac{1}{12}x^2+O(x^3)$$

Hence $1+n\log \frac{2n-1}{2n+1}=-\frac{1}{12n^2}+O(\frac 1{n^3})$

And your series is convergent.

Jean-Claude Arbaut
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