Convergence of $\sum_{k=1}^n(1-k/n)a_k$ Assume that the series $\displaystyle \sum_{n=1}^\infty a_n$ converges to a finite number, say $S$. Now let's consider a sequence of modified partial sums $\displaystyle S_n=\sum_{k=1}^n(1-\frac{k}{n})a_k$. 
It is easy to see that, if $\displaystyle \sum_{n=1}^\infty |a_n|<\infty$, then $\displaystyle \lim_{n\to \infty} S_n=S$.
My questions is, could we relax the condition on the absolute convergence? Does $S_n$ always converge to $S$ (whenever $\sum a_n$ converges)?
Thanks!

No additional conditions other than the convergence of the original series are needed.
Since the sum of $a_k$ converges, for any $\epsilon\gt0$, there is an $N$ so that if $m,n\ge N$
$$
\left|\,\sum_{k=m}^na_k\,\right|\le\epsilon/2
$$
Therefore, for $m,n\ge N$,
$$
\begin{align}
\left|\,\frac1n\sum_{k=m}^nka_k\,\right|
&=\left|\,\frac1n\sum_{k=m}^n\sum_{j=1}^k1\cdot a_k\,\right|\\
&=\left|\,\frac1n\sum_{j=1}^n\sum_{k=\max(j,m)}^na_k\,\right|\\
&\le\frac1n\sum_{j=1}^n\epsilon/2\\[9pt]
&=\epsilon/2
\end{align}
$$
For $\displaystyle n\ge\frac2\epsilon\sum\limits_{k=1}^{N-1}ka_k$, we then have
$$
\begin{align}
\left|\,\frac1n\sum_{k=1}^nka_k\,\right|
&\le\left|\,\frac1n\sum_{k=1}^{N-1}ka_k\,\right|
+\left|\,\frac1n\sum_{k=N}^nka_k\,\right|\\[9pt]
&\le\epsilon/2+\epsilon/2\\[16pt]
&=\epsilon
\end{align}
$$
Since $\epsilon$ was arbitrary,
$$
\lim_{n\to\infty}\frac1n\sum_{k=1}^nka_k=0
$$
and therefore,
$$
\begin{align}
\lim_{n\to\infty}\frac1n\sum_{k=1}^n(n-k)a_k
&=\lim_{n\to\infty}\sum_{k=1}^na_k-\lim_{n\to\infty}\frac1n\sum_{k=1}^nka_k\\
&=\sum_{k=1}^\infty a_k
\end{align}
$$
