Show that if $\sum \limits_{n=1}^\infty a_n$ converges then $\lim\limits_{n\to\infty}\frac{1}{n}\sum _{k=1}^n ka_k=0$ Show that if $\sum \limits_{n=1}^\infty a_n$ converges then $\lim\limits_{n\to\infty}\frac{1}{n}\sum \limits_{k=1}^n ka_k=0$
since given that  $\sum\limits _{n=1}^\infty a_n$ converges 
then we can say that $\lim _{n\to \infty} a_n=0$
but i don't get any idea to prove this result how to solve this problem

$$\sum \limits_{n=1}^\infty a_n<\infty$$ converges then
for $\varepsilon >0$ there is $n_0$ such that 
$$\left|\sum \limits_{n=n_0}^\infty a_n\right|<\varepsilon$$
let $n\ge n_0$ then, 
$$\left|\frac{1}{n}\sum \limits_{k=1}^{n}ka_k\right| \le\left|\frac{1}{n}\sum \limits_{k=1}^{n_0} ka_k\right|+\left|\frac{1}{n}\sum \limits_{k=n_0+1}^n ka_k\right| \le \frac{M}{n}+\left|\sum \limits_{k=n_0+1}^\infty a_k\right| \le\frac{M}{n} +\varepsilon\to 0 $$
 Where $M= |\sum \limits_{k=1}^{n_0} ka_k|$ and one note that in the second term $k<n$ for all $k\in\{n_0+1,\cdots, n\}$ which yields
$$\left|\frac{1}{n}\sum \limits_{k=n_0+1}^n ka_k\right|\le \left|\sum \limits_{k=n_0}^\infty a_k\right|\le \varepsilon $$
