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This question already has an answer here:

How do I find de sum of this serie using the partial sums?

$$\sum_{n=1}^\infty \log\left[\frac{(n+1)^2}{n(n+2)}\right]$$

Thanks!

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marked as duplicate by user147263, Pragabhava, colormegone, user296602, Winther Jan 25 '16 at 2:29

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  • $\begingroup$ The sum of the logarithms is the logarithm of the product. Telescope $\endgroup$ – Henry Jan 25 '16 at 1:12
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For $k$ fixed:

$\begin{eqnarray} \sum_{n=1}^k\log\left(\frac{(n+1)^2}{n(n+2)}\right)&=&\sum_{n=1}^k2\log(n+1)-\log(n)-\log(n+2)\\ &=&\left(\sum_{n=1}^k\log(n+1)-\log(n)\right)+\left(\sum_{n=1}^k\log(n+1)-\log(n+2)\right)\\ &=&(\log(k+1)-\log(1))-(\log(k+2)-\log(2))\\ &=&\log\left(\frac{2(k+1)}{k+2}\right)\\ &\to&\log2 \end{eqnarray}$

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