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$\sum_{n=1}^{+\infty}\left[n \log\left ( \frac{2n+1}{2n-1} \right )-1\right]$ ,converges or not

My attempt

$a_n=n \log\left ( \frac{2n+1}{2n-1} \right )-1 \\ \\ \log \left ( \frac{2n+1}{2n-1} \right )=\log \left ( 1+\frac{2}{2n-1} \right ) \sim \frac{2}{2n-1} \ (n\rightarrow +\infty) \\ \\ a_n=n\log \left ( 1+\frac{2}{2n-1} \right )-1 \sim n \frac{2}{2n-1}-1=\frac{2n}{2n-1}-1=\frac{1}{2n-1} \\ \\ $

$\sum_{n=1}^{+\infty}\frac{1}{2n-1} $ diverges ,then $\sum_{n=1}^{+\infty}a_n $ diverges

but according to Wolfram $\sum_{n=1}^{+\infty}a_n=\frac{1}{2}-\frac{\log 2}{2}$

where is the mistake in my solution please?

thanks

Felix Marin
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Student
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2 Answers2

2

$$\log\left(\frac{2n+1}{2n-1}\right) = 2\,\text{arctanh}\left(\frac{1}{2n}\right)=\frac{1}{n}+\frac{1}{12n^3}+O(n^{-5})$$ leads to $$ n\log\left(\frac{2n+1}{2n-1}\right)-1 = \frac{1}{12n^2}+O(n^{-4}) $$ so the given series is convergent. Your initial approximation was simply not accurate enough.

The exact value of the series is simple to compute through summation by parts. We have: $$\begin{eqnarray*} \sum_{n=1}^{N}\left[n\log\left(\frac{2n+1}{2n-1}\right)-1\right]&=&-N+\sum_{n=1}^{N}n\left[\log(2n+1)-\log(2n-1)\right]\\&=&-N+N\log(2N+1)-\sum_{n=1}^{N-1}\log(2n+1)\\&=&-N+N\log(2N+1)-\log((2N-1)!!)\end{eqnarray*}$$ so $$ \sum_{n=1}^{+\infty}\left[n\log\left(\frac{2n+1}{2n-1}\right)-1\right] = \frac{1-\log 2}{2}$$ follows from Stirling's inequality. Have a look at this recent question of mine.

Community
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Jack D'Aurizio
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2

Your mistake was assuming this: If $x_n \sim y_n,$ then $x_n - 1 \sim y_n -1.$ A counterexample is $x_n =1+ 1/n, y_n = 1+1/n^2.$

zhw.
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