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I don't know why but I'm having a hard time determining whether this series $$ \sum\limits _{n=1}^{\infty}\ln\left(\frac{\left(n+1\right)^{2}}{n\left(n+2\right)}\right) $$ converges to a real limit.

I did try to break it down according to $\ln$ identities. $$ \ln\left(\frac{\left(n+1\right)^{2}}{n\left(n+2\right)}\right)=\ln\left(\left(n+1\right)^{2}\right)-\ln\left(n\left(n+2\right)\right)=2\ln\left(n+1\right)-\ln n-\ln\left(n+2\right) $$ and then tried to increase it to get to a series that converges to a real number: $$ 2\ln\left(n+1\right)-\ln n-\ln\left(n+2\right)\leq 2\ln\left(n+1\right)-\ln n-\ln\left(n+1\right)=\ln\left(n+1\right)-\ln n $$ so $S_n$ will be like that $$ S_n=\ln2-\ln1+\ln3-\ln2+\dots+\ln(n+1)-\ln n \\=\ln (n+1)-\ln1=\ln (n+1) $$ and so $\lim_{n\to\infty}S_n=\infty$

I know this series does converge to a real number (Well according to Wolfram Alpha :) $\ $)

Any help would be appreciated.

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up vote 2 down vote accepted


Theorem: Let $\lim_{n\to\infty}n^pu_n=A$. $$\sum u_n$$ converges if $p>1$ and $A$ is finite.

Now take $p=2$.

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Babak S:▄▀▄▀▄▒▒N▒i▒c▒e▒▒h▒i▒n▒t▒▒▄▀▄▀▄ – Software May 11 '13 at 14:48
Nice work, Babak! Nice work, you too, @flashdesign2550 ;-) – amWhy May 12 '13 at 0:26
Thank you(◕ ‿-) ,✿.*,✿.*,✿.* ,✿ ,✿.*,✿.*,✿.,✿.*,✿.*,✿.*’,✿.*✿, ``,✿.*,✿.*,✿.* ,✿.*,°*’.*,✿.*✿ ,✿.*,✿.*,✿.,✿.*,✿.*,✿.*’,✿.*,✿ ``,✿.*,✿.*,✿.* ,✿.*,°*’ .*,✿.*✿ ,✿.*,✿.*,✿.,✿.*,✿.*,✿.*’,✿.*`,✿ – Software May 12 '13 at 9:46

Hint:$\dfrac{(n+1)^2}{n(n+2)}=1+\dfrac{1}{n(n+2)}$ and $\log(1+x)=x+O(x^2)$.

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