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Determine whether $\sum_{n=8}^\infty$ $\ln\left(1 + \frac{1}{n^2-1} \right)$ converges and find the sum if so.

So far..
$\ln\left(1 + \frac{1}{n^2-1}\right) = ln\left(\frac{n^2}{n^2-1}\right)$
This seems to diverge by the nth term test but I have a feeling that it should converge and I am doing something wrong here...

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  • $\begingroup$ The $n$th term test would be (by your algebra): $\displaystyle\lim_{n \to \infty} \ln\left(\dfrac{n^2}{n^2+1}\right)=\ln \left(\lim_{n \to \infty} \dfrac{n^2}{n^2-1}\right)=\ln 1=0$ so that it is inconclusive. $\endgroup$ – mathematics2x2life Oct 19 '16 at 17:38
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$$\ln(1+\frac{1}{n^2-1})=\ln\frac{n^2}{n^2-1}=2\ln n-\ln(n+1)-\ln(n-1)$$

$$\sum_{n=8}^k \ln(1+\frac{1}{n^2-1})=2\sum_{n=8}^k\ln n-\sum_{n=9}^{k+1}\ln n-\sum_{n=7}^{k-1}\ln n=\ln k-\ln(k+1)-\ln7+\ln 8$$

Let $S_k=\sum_{n=8}^k\ln(1+\frac{1}{n^2-1}).$

We then have $S_k=\ln(\frac{k}{k+1})+\ln(8/7)<\ln(1)+\ln(8/7)$. We have the sequence $(S_k)_{k=8}^{\infty}$bounded.

And $S_{k+1}-S_k=\ln(1+\frac{1}{(k+1)^2-1})>\ln1=0$. So $(S_k)$ is increasing. Thus by the Monotone Convergence Theorem, $S_k$ must converge, i.e. $\sum_{n=8}^\infty\ln(1+\frac{1}{n^2-1})$ exists and is finite.

Take $k\to \infty $, the series converge to $\ln(8/7)$

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  • $\begingroup$ This is very helpful thank you! Can you please explain how you would check for convergence? Is it sufficient to say that $ln(1+\frac{1}{n^2-1}) < \frac{1}{n^2-1}$ and that since $\frac{1}{n^2-1}$ converges by p-series, then $ln(1+\frac{1}{n^2-1})$ converges by Direct Comparison Test? Or is more rigorous proof needed? $\endgroup$ – Tatiana Frank Oct 20 '16 at 3:06
  • $\begingroup$ added a proof of convergence in terms of sequence, but your proof is also correct. @TatianaFrank $\endgroup$ – Nick Oct 20 '16 at 6:07
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You can try to prove : $\sum_{n=8}^i\ln\left(1 + \frac{1}{n^2-1} \right)=\ln\left(\prod_{n=8}^i\left(\frac{n^2}{(n+1)(n-1)}\right)\right)=\ln\left(\frac{8i}{7(i+1)}\right)$

Hint : It is a telescopic product.

Then you should be able to conclude.

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  • $\begingroup$ I have not learned $\prod$ yet... I'm sure there is a way to solve without it... $\endgroup$ – Tatiana Frank Oct 19 '16 at 17:26
  • $\begingroup$ You can write the equality without this symbol : $\ln(\frac{8^2}{7\times9}\frac{9^2}{8\times10}\dots\frac{(i-1)^2}{(i-2)i}\frac{i^2}{(i-1)(i+1)})$. Then you can delete all the terms that appear in both the numerator and the denomntator, and you will have the same result. $\endgroup$ – Jennifer Oct 19 '16 at 17:41
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Hint: $\ln (1+x) < x$ for positive $x.$

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  • $\begingroup$ are you saying I should use comparison test and compare to just x?... $\endgroup$ – Tatiana Frank Oct 19 '16 at 17:28
  • $\begingroup$ Yes, where $x = 1/(n^2-1).$ (This only proves convergence, not the actual value of the sum.) $\endgroup$ – zhw. Oct 19 '16 at 17:36

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