# Show that $\sum_{n=1}^{\infty}\ln(1+a_n)$ converges.

Show that if $$a_n>0$$ and $$\sum_{n=1}^\infty a_n$$ converges, then $$\sum_{n=1}^{\infty}\ln(1+a_n)$$ converges too.

My attempt:

Since $$a_n>0$$ and $$\sum_{n=1}^\infty a_n$$ converges, $$\lim_{n\to\infty} a_n$$ must be $$0$$.

If we set $$a_n=\ln(1+a_n)$$ and $$b_n = a_n$$ and use the limit comparison test we have

$$\lim_{n\to\infty} \frac{\ln(1+a_n)}{a_n} =\lim_{x\to\infty}\frac{\ln(1+x)}{x} =\lim_{x\to\infty}\frac{1/(1+x)}{1}=0$$

but when the limit is $$0$$ we can't conclude anything about the convergence of the series, what was my mistake here?

• Can you show that when $a_n>0$ you have $0 \lt \log(1+a_n) \le a_n$? Jan 18, 2023 at 15:19
• This step doesn't look right to me: $\lim_{n\to\infty} \frac{\ln(1+a_n)}{a_n}=\lim_{x\to\infty}\frac{\ln(1+x)}{x}$ Jan 18, 2023 at 15:21

Using the following property :

Under assumptions :

1. If $$a_n$$ is positive sequence
2. If $$\sum_{\infty} a_n$$ converges
3. If a sequence $$b_n$$ is such $$a_n \sim_{\infty} b_n$$

So :

$$\sum_{\infty} b_n \ \ \text{converges}$$

Since $$a_n>0$$, $$a_n \to 0$$ because $$\sum_{\infty} a_n$$ converging. $$0

You have you result.

When you change the limit from being in terms of $$n$$ to $$x$$ you have to consider that because $$a_n$$ goes to 0, $$x$$ has to go to 0, so the limit is when $$x\rightarrow 0$$ not $$+\infty$$, and with this you get the famous limit of $$\frac{\log (1+x)}{x}$$ that is equal to 1.