I'm having trouble understanding if the following series converges or diverges:

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

I've noticed that $\lim _{x\to \infty \:}\ln \left(\frac{x}{x+1}\right) = 0$ and therefore I can't deduce that it diverges, but other than that I'm really not sure what is the right way to go here. Any help is appreciated

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    $\begingroup$ $\log(\frac{n}{n+1}) = \log(n) - \log(n+1)$ may be of use. $\endgroup$ – Countingstuff May 5 '18 at 20:59

This is a telescoping sum $$ \sum _{n=1}^{\infty }\ln\left(\frac{n}{n+1}\right)=\ln(1)-\ln(2)+\ln(2)-\ln(3)+\ln(3)-\ln(4)+... $$

This means that the first condition of sum having at least one accumulation point is not fulfilled.

$$\sum _{n=1}^{m }\ln\left(\frac{n}{n+1}\right)=-\ln(m+1)$$

For this reason the series diverges.



Note that

$$\ln\left(\frac{n}{n+1}\right)=\ln\left(1-\frac{1}{n+1}\right)\sim-\frac 1{n+1}$$

then refer to limit comparison test.

  • $\begingroup$ Equivalents (in the Landau sense, like this) by definition mean that we want the first non-zero order term. You write $\sim \frac{1}{n+1}$ instead of $\sim\frac{1}{n}$: this is technically correct, but misleading. $\sim \frac{1}{n + n^{0.999999}}$ would be equally correct. $\endgroup$ – Clement C. May 5 '18 at 21:57
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    $\begingroup$ @ClementC. Yes you are right. I suggested that keeping in mind the limit comparison test with $\frac{\ln\left(1-\frac{1}{n+1}\right)}{\frac 1{n+1}}$ but of course also $\frac1n$ is fine. Thanks! $\endgroup$ – user May 5 '18 at 22:04

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