How to prove a series is convergent?

Question

For this proof, I'm having trouble completing it. Here is what I have so far. Can anyone please help me out?

Show that if $a_n > 0$ for all n ∈ N and $\sum a_n$ is convergent, then $\sum ln(1+a_n)$ is convergent.

Suppose $\sum a_n$ is convergent

$\sum a_n$

$= a_1+a_2+...+a_n$

$ = S_n$

$\lim_{n\to \infty} S_n = L$

$\sum ln(1+a_n)$

$ = ln(1+a_1+a_2+...+a_n)$

$ = ln(1+S_n)$

Answer 1

Hint: use the limit comparison test. In order to show $\sum \ln(1+a_n)$ converges, you just need to show that $$ \lim_{n\to\infty} \frac{\log(1+a_n)}{a_n} $$ exists and is nonzero and finite. Since $a_n\to0$ as $n\to\infty$, this limit equals $$ \lim_{x\to0 } \frac{\log(1+x)}{x} $$ which can be computed using L'Hôpital's rule.

Answer 2

Using the Taylor series, you can check easily than $$0\le x < 1\implies\log(1 + x)\le x,$$ so $$0\le\log(1 + a_n)\le a_n$$ and now apply comparison test.

Answer 3

Since, for $x \ge 0$, $\ln(1+x) =\int_0^x \dfrac{dt}{1+t} $, we have $\ln(1+x) \le \int_0^x \dfrac{dt}{1} =x $ and $\ln(1+x) \ge\int_0^x \dfrac{dt}{1+x} =\dfrac{x}{1+x} $.

Therefore $\sum \ln(1+a_n) \le \sum a_n $.

For the other way, if $a_n \to 0$ and $a_n > 0$, there is a $N$ such that $a_n < 1$ for $n \ge N$.

Therefore if $\sum \ln(1+a_n)$ converges then $\sum_{n \ge N} \ln(1+a_n) \ge \sum_{n \ge N} \dfrac{a_n}{1+a_n} \ge \sum_{n \ge N} \dfrac{a_n}{2} $ so $\sum a_n$ converges.