Given: $\sum n a_n$ is convergent. To prove: The sequence ${a_n}$ converges Given: $\sum n a_n$ is convergent. To prove: The sequence ${a_n}$ converges.
My proof: 
Since $\sum n a_n$ is convergent, so $na_n \rightarrow 0,$
i.e., $\frac{a_n}{\frac{1}{n}}\rightarrow 0. $
As the denominator of tends to $0$, so the limit $\frac{a_n}{\frac{1}{n}}$ would exist only if the numerator ${a_n}$ tends to $0$, so that an indeterminate form is formed. Thus ${a_n}$ is convergent.
Is it okay? I am not satisfied with it. 
It seems to me that the proof lacks some explanation. 
I think the argument 
"As the denominator of tends to $0$, so the limit $\frac{a_n}{\frac{1}{n}}$ would exist only if the numerator tends to $0$" 
requires some more explanation.

For $n\neq 0$, $0\leq |a_n|\leq |na_n|$. Use the squeeze theorem. 
