# If $a_n\geq 0$ and $\sum_{n=1}^{\infty} a_n$ is convergent, then $\sum_{n=1}^{\infty} a_n^2$ and $\sum_{n=1}^{\infty} \frac{a_n}{a_n+1}$ converges

If $a_n\geq 0$ and $\sum_{n=1}^{\infty} a_n$ convergent, then $\sum_{n=1}^{\infty} a_n^2$ and $\sum_{n=1}^{\infty} \frac{a_n}{a_n+1}$ converges

For $\sum a_n^2$ I used this to prove it like this: since $\sum_{n=1}^{\infty}a_n$ converges, then $\lim a_n \to 0$, so $|a_n| <\epsilon$ for large $n$. Choosing $\epsilon=1$ we have $|a_n|<1\implies a_n^2 < |a_n|<1$. By comparsion, $a_n^2 < a_n$ for sufficiently large $n$, then $\sum a_n^2$ converges.

For $\sum \frac{a_n}{a_n+1}$ I tried but couldn't find a comparsion for $a_n$ and $\frac{a_n}{a_n+1}$. I tried using that since $\sum a_n$ converges, then $|a_n|<1$ for large $n$. Then $a_n+1<a_n+a_n = 2a_n\implies \frac{1}{a_n+1}<\frac{1}{2a_n}\implies \frac{a_n}{a_n+1}<\frac{a_n}{2a_n}$.

Ok, that didn't help

hint: $\dfrac{a_n}{a_n+1} \leq a_n$, and $a_n < 1, \forall n \geq n_{0}\implies a_n^2 \leq a_n, n \geq n_{0}$