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Is this infinite series convergent or not? How to solve it?

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

Context: I've learnt infinite series for a couple of weeks but still not familiar with how to solve those questions. I've tried to use Taylor to transform it but still no result.

Can anybody help?

Thanks a lot!

Mostafa Ayaz
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berryx
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  • Hi, welcome to Math SE. Hint: using $x=\frac{1}{2n}$ in $\frac{1}{2x}\ln\frac{1+x}{1-x}=1+\frac13x^2+\cdots$ will get you an asymptotic form for the $n$th term. – J.G. Sep 13 '23 at 14:56
  • thanks!I think I get how to solve it in general.but I'm not quite sure about your second equation.maybe some Taylor stuff that I haven't get yet? – berryx Sep 13 '23 at 15:04
  • Use $\ln(1+x)=x-\frac12x^2+\frac13x^3-\cdots$ for $|x|<1$ twice in $\ln(1+x)-\ln(1-x)$. – J.G. Sep 13 '23 at 15:06
  • ohhhh amazing interesting!but the next we may add an infinite number of convergent series.is that still converges?since we have infinite ones. or can I just write it into the form of o(..)?is that reasonable? – berryx Sep 13 '23 at 15:13
  • That the $n$th term is $O(1/n^2)$ is enough to use a comparison test. – J.G. Sep 13 '23 at 15:18
  • ..almost forgot. got it. thank you! – berryx Sep 13 '23 at 15:28
  • https://math.stackexchange.com/questions/3019050/find-sum-of-series-sum-n-1-infty-n-cdot-ln-frac2n12n-1-1?noredirect=1 – Svyatoslav Sep 13 '23 at 16:03

3 Answers3

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You can use limit comparison test with $\sum \frac{1}{n^2}$, with some work (L'Hopital rule or manipulations) you get $$ \lim_{n \to +\infty} \frac{n\ln{\left(\frac{2n+1}{2n-1}\right)}-1}{\frac{1}{n^2}} = \frac{1}{12}, $$ since $\sum \frac{1}{n^2}$ converges, the series in question will also converge.

Raul Fernandes Horta
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Using $\frac{2n+1}{2n-1}=1+\frac2{2n-1}$ and $x-\frac{x^2}2<\log(1+x)<x-\frac {x^2}2+\frac{x^3}3$ for $0<x<1,$ then using $x=\frac2{2n-1}$ then $$x-\frac{x^2}2=\frac{2}{2n-1}-\frac{2}{(2n-1)^2}=\frac{4n-4}{(2n-1)^2}=\frac1n\left(1-\frac1{(2n-1)^2}\right)$$ and thus:

$$-\frac{1}{(2n-1)^2}<n\log\frac{2n+1}{2n-1}-1<\frac{2n+3}{3(2n-1)^3}<\frac{1}{(2n-1)^2}$$ when $n\geq 2$ and thus $2n+3<3(2n-1).$

Thomas Andrews
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(Direct) Comparison test: $$n\ln\left(\frac{2n+1}{2n-1}\right)-1<\frac1{n^2}$$ $$2n+1<e^{(\frac1n+\frac1{n^3})}(2n-1)$$ It is enough to show that $$2n+1<(1+\frac1n+\frac1{n^3}+\frac1{2n^2})(2n-1)$$ which is true when $n>\frac23.$

Bob Dobbs
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