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$\require{cancel}$

Evaluate $$\sum_{n=2}^{\infty}\ln\left(\frac{n^2-1}{n^2}\right)$$ My procedure: $$\sum_{n=2}^{\infty}\ln\left(\frac{n^2-1}{n^2}\right)=\sum_{n=2}^{\infty}(\ln\left(n-1\right)-\ln(n)+\ln(n+1)-\ln(n))$$ If we evaluate a few terms: $$\sum_{n=2}^{m}(\ln\left(n-1\right)-\ln(n)+\ln(n+1)-\ln(n))= $$ $$=0-\ln(2)\cancel{+\ln(3)}\cancel{-\ln(2)+\ln(2)}\cancel{-\ln(3)}\cancel{+\ln(4)} \cancel{-\ln(3)+\ln(3)}\cancel{-\ln(4)}+\ln(5)-\ln(4)+...+\ln(m+1)-\ln(m)$$ Thus: $$\lim_{m\to\infty}\sum_{n=2}^{m}\ln\left(\frac{n^2-1}{n^2}\right)=\lim_{m\to\infty}\left(-\ln(2)+\ln(m+1)-\ln(m)\right)=\boxed{-\ln(2)}$$ However, I am not really happy with this method. Is there any other way? A more elegant one?

DMH16
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    What do you not like about this method of telescoping sum? – Jacky Chong Mar 07 '17 at 02:06
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    That's a perfectly fine exercise in telescoping. Are you maybe asking how it can be "formalized" beyond just crossing out the terms "by inspection"? – dxiv Mar 07 '17 at 02:12
  • @JackyChong The fact that I am not formalizing the telescoping series – DMH16 Mar 07 '17 at 02:13
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    You can consider

    $$\prod^\infty_{n=2} \frac{n^2-1}{n^2} $$

    But I think your approach is the best to tackle this problem.

     – Zaid Alyafeai Mar 07 '17 at 02:13
  • @dxiv yes that's it – DMH16 Mar 07 '17 at 02:13
  • It might be worth noticing that the parial sums can be handily calculated algebraically with the factorial function (instead of making inductive reasonings) $$\sum_{k=2}^m\ln\frac{(n+1)(n-1)}{n^2}=\ln\frac{\left(\prod_{k=2}^m(k+1)\right)\left(\prod_{k=2}^m(k-1)\right)}{\left(\prod_{k=2}^mk\right)^2}=\=\ln\frac{\frac{(m+1)!}2(m-1)!}{m!^2}=\ln\frac{m+1}{2m}$$ That being said, it's just a reharsal of the same idea. –  Mar 07 '17 at 02:20

2 Answers2

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Step by step telescoping below:

$$\require{cancel} \begin{align} & \sum_{n=2}^{\infty}\big(\ln\left(n-1\right)-2 \ln(n)+\ln(n+1)\big) \\ &\;\;= \sum_{n=2}^{\infty}\big(\left(\ln(n+1)-\ln(n)\right)-\left(\ln(n)-\ln\left(n-1\right)\right)\big) \\ &\;\;= \sum_{n=2}^{\infty}\big(\ln(n+1)-\ln(n)\big)- \sum_{n=2}^{\infty}\big( \ln(n)-\ln\left(n-1\right)\big) \\ &\;\;= \sum_{\color{red}{n=3}}^{\infty}\big(\ln(n)-\ln(n-1)\big)- \sum_{n=2}^{\infty}\big( \ln(n)-\ln\left(n-1\right)\big) \\ &\;\;= \bcancel{\sum_{n=3}^{\infty}\big(\ln(n)-\ln(n-1)\big)}-\big(\ln(2)-\ln(2-1)\big)- \bcancel{\sum_{\color{red}{n=3}}^{\infty}\big( \ln(n)-\ln\left(n-1\right)\big)} \\[3px] &\;\;= -\ln(2) \end{align} $$

dxiv
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    Perfect! This is what I was looking for! – DMH16 Mar 07 '17 at 02:25
  • @DMH16 Glad it helped. I should add that the above was meant to show the mechanism of telescoping, but it does not in and by itself prove the convergence of the given series, since it works on the entire series, and the canceling out at the end pre-supposes that both infinite sums converge. – dxiv Mar 07 '17 at 03:46
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    To actually prove convergence, you'd have to work out the partial sums, but the mechanism is very similar:

    $$\require{cancel} \sum_{n=2}^m (a_{n+1}-a_{n})=\sum_{n=2}^m a_{n+1}-\sum_{n=2}^m a_{n}=\sum_{\color{red}{n=3}}^{\color{red}{m+1}} a_{\color{red}{n}}-\sum_{n=2}^m a_{n}=\bcancel{\sum_{n=3}^{\color{red}{m}} a_{n}}+a_{m+1}-a_2-\bcancel{\sum_{\color{red}{n=3}}^m a_{n}} $$

    With $a_n=\ln(n)-\ln(n-1)$ this reduces to the partial sum formula you derived in the original post.

     – dxiv Mar 07 '17 at 03:46
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Another way would be to consider the exponentiation. Observe \begin{align} \exp\left(\sum^N_{n=2}\ln\frac{n^2-1}{n^2} \right)=&\ \prod^N_{n=2}\frac{n^2-1}{n^2} = \prod^N_{n=2}\frac{(n-1)(n+1)}{n \cdot n}\\ =&\ \frac{(2-1)(2+1)}{2\cdot 2}\cdot \frac{(3-1)(3+1)}{3\cdot 3}\cdots\frac{(N-2)N}{(N-1)\cdot (N-1)}\frac{(N-1)(N+1)}{N\cdot N}\\ =&\ \frac{1}{2}\frac{N+1}{N} \end{align} which means \begin{align} \sum^N_{n=2}\ln \frac{n^2-1}{n^2} = \ln\frac{N+1}{2N}. \end{align} Take the limit as $N\rightarrow \infty$ yields $-\ln 2$.

Jacky Chong
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