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How would I find the formula for the partial sum $S_n $ of this series? $$\sum_{k=1}^\infty \ln\frac{k(k+2)}{(k+1)^2}$$

I know that $S_1=\ln(3/4) $, $S_2=\ln(2/3) $, $S_3 =\ln(5/8)$, and $S_4=\ln(3/5)$, but I don't see any pattern that would allow me to write the partial sum formula.

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hint: $\ln\left(\dfrac{k(k+2)}{(k+1)^2}\right) = (\ln(k) - \ln(k+1)) + (\ln(k+2) - \ln(k+1))$

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  • $\begingroup$ is $S_n=\ln(1)-\ln(2)+\ln(n+2)-\ln(n+1)$? $\endgroup$ – user532874 Oct 16 at 1:38
  • $\begingroup$ In which case the original series diverges right? $\endgroup$ – user532874 Oct 16 at 1:40
  • $\begingroup$ are you still there deep sea $\endgroup$ – user532874 Oct 16 at 1:58
  • $\begingroup$ Yes I am. I will edit it for u soon.😀 $\endgroup$ – DeepSea Oct 16 at 8:48
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Consider the partial sum, $$S_{n} = \sum_{k=1}^{n} \ln\left(\frac{k \, (k+2)}{(k+1)^{2}}\right) $$ which can be evaluated in two ways. The first being: \begin{align} S_{n} &= \sum_{k=1}^{n} \ln\left(\frac{k \, (k+2)}{(k+1)^{2}}\right) \\ &= \sum_{k=1}^{n} ( \ln(k) + \ln(k+2) - 2 \, \ln(k+1) ) \\ &= \sum_{k=2}^{n} \ln(k) + \sum_{k=3}^{n+2} \ln(k) - 2 \, \sum_{k=2}^{n+1} \ln(k) \\ &= \sum_{k=3}^{n+2} \ln(k) - \sum_{k=2}^{n+1} \ln(k) - \ln(n+1) \\ &= \ln(n+2) - \ln(n+1) - \ln(2) = \ln\left(\frac{n+2}{2 \, (n+1)} \right). \end{align}

The second is: \begin{align} S_{n} &= \sum_{k=1}^{n} \ln\left(\frac{k \, (k+2)}{(k+1)^2}\right) = \ln\left(\prod_{k=1}^{n} \frac{k(k+2)}{(k+1)^2}\right) \\ &= \ln\left(\frac{n! \, \prod_{k=3}^{n+2} k}{\prod_{k=2}^{n+1} k^2} \right) \\ &= \ln\left(\frac{n! \, (n+2)!}{2! \, ((n+1)!)^2}\right) = \ln\left(\frac{n! \, (n+2)}{2 \, (n+1)!} \right) \\ &= \ln\left(\frac{n+2}{2(n+1)}\right). \end{align}

In both cases $$S_{n} = \ln\left(\frac{1 + \frac{2}{n}}{2 \, \left(1 + \frac{1}{n}\right)} \right)$$ such that as $n \to \infty$ the result becomes $- \ln(2)$ which yields $$\sum_{k=1}^{\infty} \ln\left(\frac{k \, (k+2)}{(k+1)^{2}}\right) = - \ln(2).$$

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Hint

$$\sum_{k=1}^N \ln\frac{k(k+2)}{(k+1)^2} = \ln \left(\prod_{k=1}^N \frac{k(k+2)}{(k+1)^2} \right)= \ln \left(\frac{(\prod_{k=1}^N k)(\prod_{k=1}^N (k+2))}{(\prod_{k=1}^N (k+1))^2} \right)= \ln \left(\frac{N! \frac{(N+2)!}{2}}{((N+1)!)^2} \right)$$

Cancel.

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