Show that if $\sum a_n$ converges absolutely, then $\sum (\tfrac{n+1}{n})a_n$ also converges absolutely.

Question

Prove that if $\sum_{n=1}^\infty a_n$ is convergent, then the series $$\sum_{n=1}^\infty \left(\frac{n+1}{n}\right)a_n$$ is also convergent.

Edit: Originally, I thought about doing a limit comparison test, where your $b_n$ would get $\frac{1}{n}$. Then that limit of $\left(\frac{n(n+1)}{n^2}\right)$ would be 1. So the series turns is comparable to $a_n$. But it seems wrong to ignore $a_n$ when doing the limit comparison.

Edit 2: My mind doesn't function so the attempt of the above edit isn't good for the problem.

Answer 1

Because

$$ \sum \left|(\frac{n+1}{n} ) a_n \right| = \sum \left| \left(\frac{1}{n} + 1\right) a_n\right| < \sum |a_n| + \sum |a_n|$$

both of which converge.

Answer 2

ASssuming $a_n > 0$, you can do the following.

Comparison Test says if you have two series, $\sum a_n$ and $\sum b_n$, and $b_n < a_n$ for all $n$, then $\sum a_n < \infty \implies \sum b_n < \infty$.

It is easy to notice that for $n > 2$, we have $(n+1)/n^2 < 1$ hence $a_n (n+1)/n^2 < a_n$...