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

Question

I am studying the convergence of $\sum _{n=1}^{\infty } \ln \left(\frac{n+1}{n}\right)$.

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Approach 1: Limit comparison test

Using the limit comparison test, and choosing $a_n = \ln \left(\frac{n+1}{n}\right)$ and $b_n = \frac1n$ it is obvious that the sum diverges:

$\lim _{n\to \infty \:}\left(\frac{\ln \left(\frac{n+1}{n}\right)}{\frac{1}{n}}\right) = 1$

But I used De L'Hospital's theorem to evaluate this limit by letting $f(x)$ instead of $f(n)$, hence I decided to evaluate the sum in a more sequence-centric way.

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Approach 2: Telescoping Series

$\sum _{n=1}^{\infty } \ln \left(\frac{n+1}{n}\right) = \sum _{n=1}^{\infty } \ln(n+1) - \ln(n) = \lim_{n\to\infty} -\ln(1) + \ln(n) = +\infty \text { (Diverges) }$

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Although the result of the second approach is the same, I am not sure about it. Is it correct?

Answer 1

In the first case we don’t need L’Hospital, simply note that

$$\frac{\ln\left(1+\frac{1}{n}\right)}{\frac1n}= \ln\left(1+\frac{1}{n}\right)^n\to \ln e=1$$

The second approach is fine.

Answer 2

Hint

$$\ln\left(1+\frac{1}{n}\right)\sim \frac{1}{n}\quad \text{when }n\to \infty.$$