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I would like to calculate the following sum:

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

I do know that it converges but I have gone that far:

\begin{align} & \sum_{n=1}^{\infty}\ln \left( \frac{n^2+2n+1}{n^2+2n} \right) \Longleftrightarrow \sum_{n=1}^\infty \ln \left( \frac{(n+1)^2}{n(n+2)} \right) \\[10pt] = {} & \ln \left( \frac{4}{3} \right)+\ln \left( \frac{9}{8} \right)+\ln \left( \frac{16}{15} \right)+\cdots+\ln \left( \frac{n}{n-1} \right)\\[10pt] = {} & \ln \left( \frac{4}{3}\frac{9}{8}\frac{16}{15} \cdots \frac{n}{(n-1)} \right)=\ln (n) \end{align} which diverges as $n\to \infty.$ It looked like telescoping in the beginning but now I am confused. Where have I gone wrong? Thanks.

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    $\begingroup$ $$\sum_{n\geq 1}-\log\left(1-\frac{1}{(n+1)^2}\right)=-\log\prod_{m\geq 2}\left(1-\frac{1}{m^2}\right)=\log 2. $$ $\endgroup$ – Jack D'Aurizio Jan 20 '17 at 0:02
  • $\begingroup$ @kingW3 But in that way, it still diverges. $\endgroup$ – Mitscaype Jan 20 '17 at 0:03
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    $\begingroup$ @kingW3 Should it be $\ln\frac{(n+1)^2}{n(n+2)}$? $\endgroup$ – Juniven Jan 20 '17 at 0:03
  • $\begingroup$ My mistake,rushed into it. $\endgroup$ – kingW3 Jan 20 '17 at 0:05
  • $\begingroup$ @JackD'Aurizio Partial fractions? For the quantity inside the logarithm $\endgroup$ – Mitscaype Jan 20 '17 at 0:05
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Rewrite the general term as $\ln\dfrac{(n+1)^2}{n(n+2)}$, use the functional property of logs and you'll obtain a telescoping product for partial sums: \begin{align}\sum_{k=1}^{n}\ln \frac{(k+1)^2}{k(k+2)}&=\ln\frac{2^2}{1\cdot 3}+\ln\frac{3^2}{2\cdot 4}+\ln\frac{4^2}{3\cdot 5}+\dotsm\dotsm\dotsm\\&\phantom{=}+\ln\frac{(n-1)^2}{(n-2)n}+\ln\frac{n^2}{(n-1)(n+1)}+\ln\frac{(n+1)^2}{n(n+2)}\\ &=\ln\frac{2^{\not2}\cdot\not 3^2\cdot4^2\dotsm\dotsm\dotsm\dotsm(n-1)^2\not n^2(n+1)^2\hspace{3em}}{1\cdot \not3\cdot\not2\cdot 4\cdot\not 3\cdot 5\dotsm(n-2)\not n(n-1)(n+1)\not n(n+2)}\\ &=\log\frac{2(n+1)}{n+2}\to \ln 2.\end{align}

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  • $\begingroup$ We thought the same, although I prefer to use $\sum$ instead of "...". $\endgroup$ – marty cohen Jan 20 '17 at 4:04
  • $\begingroup$ @marty cohen: Great minds think together! In my opinion, it's clearer to visualise how terms (or factors) cancel. That's why I preferred the product version, as it's shorter to write and read. $\endgroup$ – Bernard Jan 20 '17 at 9:44
  • $\begingroup$ Individual preferences. I like using $\sum$, and separating out the common range and remaining terms. Feels more rigorous to me. $\endgroup$ – marty cohen Jan 20 '17 at 19:23
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$\begin{array}\\ \sum_{n=1}^{m}\ln \left( \frac{n^2+2n+1}{n^2+2n} \right) &=\sum_{n=1}^{m}\ln \left( \frac{(n+1)^2}{n(n+2)} \right)\\ &=\sum_{n=1}^{m}( (2\ln(n+1)-\ln(n)-\ln(n+2))\\ &=2\sum_{n=1}^{m}\ln(n+1)-\sum_{n=1}^{m}\ln(n)-\sum_{n=1}^{m}\ln(n+2)\\ &=2\sum_{n=2}^{m+1}\ln(n)-\sum_{n=1}^{m}\ln(n)-\sum_{n=3}^{m+2}\ln(n)\\ &=2(\ln(2)+\ln(m+1)+\sum_{n=3}^{m}\ln(n))-(\ln(1)+\ln(2)+\sum_{n=3}^{m}\ln(n))-(\ln(m+1)+\ln(m+2)+\sum_{n=3}^{m}\ln(n))\\ &=2(\ln(2)+\ln(m+1))-(\ln(2))-(\ln(m+1)+\ln(m+2))\\ &=\ln(2)+\ln(m+1)-\ln(m+2)\\ &=\ln(2)+\ln(1-\frac{1}{m+2})\\ &\to \ln(2) \qquad\text{since } \ln(1-\frac{1}{m+2}) \to 0 \text{ as } m \to \infty\\ \end{array} $

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