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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!

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Indeed, if $\sum a_k$ is convergent then $\lim\limits_{n\to\infty} S_n=\sum_{k=1}^\infty a_k$.

To prove this, let $T_n=\sum_{k=1}^n a_k$, with $T_0=0$. Then, clearly $$\eqalign{ S_n&=\sum_{k=1}^n\left(1-\frac{k}{n}\right)(T_k-T_{k-1})\cr &=\sum_{k=1}^n\left(1-\frac{k}{n}\right) T_k -\sum_{k=1}^{n-1}\left(1-\frac{k+1}{n}\right) T_{k}\cr &=\sum_{k=1}^{n-1}\left(\frac{k+1}{n}-\frac{k}{n}\right)T_k= \frac{1}{n}\sum_{k=1}^{n-1}T_k } $$ Thus, by Cesàro's Lemma we conclude that $$ \lim_{n\to\infty}T_n=\ell~\Longrightarrow \lim_{n\to\infty}S_n=\ell. $$ and the desired conclusion follows.

${\bf Remark.}$ In fact, the general statement of Cesàro's Lemma allow us to show that $$ \liminf_{n\to\infty}\,T_n\leq \liminf_{n\to\infty}\,S_n\leq\limsup_{n\to\infty}\,S_n \leq \limsup_{n\to\infty}\,T_n.$$ Thus, if $\sum a_k=+\infty$ we have also $\lim_{n\to\infty}S_n=+\infty$.

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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, $$ \begin{align} \lim_{n\to\infty}\frac1n\sum_{k=1}^n(n-k)a_k &=\lim_{n\to\infty}\sum_{k=1}^na_k-\lim_{n\to\infty}\frac1n\sum_{k=1}^nka_k\\ &=\sum_{k=1}^\infty a_k \end{align} $$

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  • $\begingroup$ Thanks! Your proof is also very clear. $\endgroup$ – Pengfei Apr 6 '14 at 16:27

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