Prove: $(na_n)$ converges to 0 if $(a_n)_{n=1}^\infty$ is a monotonously falling sequence of non-negative real numbers Prove: Let $(a_n)_{n=1}^\infty$ be a monotonously falling sequence of non-negative real numbers, so that $\sum a_n$ is convergent. Then $(na_n)$ is a sequence converging to 0.

Well, no idea. All help is greatly appreciated!
Because $a_n$ is a non-negative real number, let's say that $c$ is the smallest number of the $a_n$ sequence. That means that $\sum_{n=1}^\infty (n *a_n) \geq \sum_{n=1}^\infty (c)$. Well, if it is that $c&gt;0$ then that series would be divergent. It is not, thus $c=0$, so the series will converge, and may only converge, to $0$.
Because $a_n$ is a non-negative real number, let's say that $c$ is the smallest number of the $a_n$ sequence. That means that $\sum_{n=1}^\infty (n *a_n) \geq \sum_{n=1}^\infty (c)$. Well, if it is that $c&gt;0$ then that series would be divergent. It is not, thus $c=0$, so the series will converge, and may only converge, to $0$.