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Let $(a_n)$ denote a sequence. Let $$S_n=\sum_{k=1}^n a_k\qquad b_n=\dfrac 1 n \sum_{k=1}^n S_k$$ Suppose that $a_n=O(\frac 1 n$) and that the sequence $(b_n)$ converges. Prove that $\sum a_n$ converges.

I'm so confused about the condition $a_n = O(\frac 1 n)$, I don't see how to apply it...

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    I don't see the relevance either. Would the general term $a_n$ be of order $\Theta(1/n)$, the series $S_n$ would diverge. – 2017-02-24
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    It is easy to see that if $\sum a_n$ converges (i.e. $S_n$ converges) then $b_n$ converges. The converse is not true in general. The condition $a_n=O(1/n)$ is called a _Tauberian_ condition, a condition under which the converse holds. – 2017-02-24
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    @pqros Any reaction to the answer below? – 2017-03-05

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This proof is wrong. See https://en.wikipedia.org/wiki/Littlewood%27s_Tauberian_theorem

Note that $b_{n}=S_{n}+\sum_{k=1}^{n}\frac{1-k}{n}a_{k}$. Since $a_{n}=O(1/n)$, for every $\epsilon >0$ exists $N$ such that $|a_{n}|\leq\frac{\epsilon}{n+N}$. So, $$ |S_{n}-b_{n}|\leq \sum_{k=1}^{n}\frac{k-1}{n}|a_{k}|\leq \frac{\epsilon}{n(n+N)}\sum_{k=1}^{n}(k-1)= o(1)+\epsilon$$ Since $b_{n}$ converges, $S_{n}$ does too.