convergence of a series in real analysis proof

Question

If $\sum_{n=1}^{\infty}a_n$ converges, prove that $\sum_{n=1}^{\infty} \dfrac{n+1}{n}a_n$ converges.

It's probably pretty trivial, but I have been staring at it for a while and cannot make any headway. Any help would be greatly appreciated

Answer 1

$\frac{n+1}{n}a_n= a_n +\dfrac{a_n}{n}$. Can you show that $\sum \dfrac{a_n}{n}$ converges?

Answer 2

Assume that $\sum_{n=1}^{\infty}a_{n}$ converges to a limit $L$. The trick to doing what you want is called summation by parts, a technique that resembles integration by parts. Let $S_{n}=\sum_{k=1}^{n}a_{k}$ for $n \ge 1$ and, for convenience, define $S_{0}=0$. Then $S_{n}-S_{n-1}=a_{n}$ for $n \ge 1$. Notice that $$ \begin{align} \sum_{n=1}^{N}\frac{1}{n}a_{n} & =\sum_{n=1}^{N}\frac{1}{n}(S_{n}-S_{n-1}) \\ & = \sum_{n=1}^{N}\frac{1}{n}S_{n}-\sum_{n=1}^{N}\frac{1}{n}S_{n-1} \\ & = \sum_{n=1}^{N}\frac{1}{n}S_{n}-\sum_{n=0}^{N-1}\frac{1}{n+1}S_{n} \\ & = \frac{S_{N}}{N}+\sum_{n=1}^{N-1}\left(\frac{1}{n}-\frac{1} {n+1}\right)S_{n}+S_{0} \\ & = \frac{S_{N}}{N}+\sum_{n=1}^{N-1}\frac{1}{n(n+1)}S_{n}. \end{align} $$ By assumption $\lim_{n\rightarrow\infty}S_{n}=L$ and, therefore, $\lim_{N\rightarrow\infty}S_{N}/N=0$. Because $S_{n}$ converges, then there is a constant $M$ such that $|S_{n}|\le M$ for all $n \ge 1$, which guarantees that $\sum_{n=1}^{\infty}S_{n}/(n(n+1))$ converges absolutely. So the sum $\sum_{n=1}^{\infty}a_{n}/n$ converges, which is what you need in order to guarantee that the following converges: $$ \sum_{n=1}^{\infty}a_{n}+\sum_{n=1}^{\infty}\frac{1}{n}a_{n}=\sum_{n=1}^{\infty}\frac{n+1}{n}a_{n}. $$

Answer 3

Given $\sum_{n=1}^{\infty}a_n = s$ is convergent, we have $$\sum_{n=1}^{\infty} \dfrac{n+1}{n}a_n = \sum_{n=1}^\infty a_n + \sum_{i=1}^\infty \dfrac{a_n}{n} = s + \sum_{i=1}^\infty \dfrac{a_n}{n}.$$ I think proving the last sum is convergent should be easy -- if not post a comment and I can give more direction.