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?

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

2 Answers 2


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. $$ 0<ln(1+a_n)\sim a_n$$

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.


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