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Assume the series $\sum u_n$ is Cesaro summable and $\lim_{n\to\infty} nu_n\to 0$. We want to see that the series is (Cauchy) convergent.

Attempt: Let $s_n=\sum_{i=1}^n u_n$ denote the $n$-th partial sum. Then $$s_{n+1}-\frac{s_1+s_2+\dots+s_n}{n}=\frac{u_2+2u_3+3u_4+\dots+nu_{n+1}}{n}$$

Since the sequence $\frac{s_1+s_2+\dots+s_n}{n}$ converges by assumption, it suffices to see that $$\lim_{n\to\infty}\frac{u_2+2u_3+3u_4+\dots+nu_{n+1}}{n}=0$$

How can we finish the proof?

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It is well known fact that if $\lim_{n\to\infty} a_n $ exists then $\lim_{n\to\infty} \frac{a_1 +a_2 +...+a_n}{n}=\lim_{n\to\infty} a_n .$ Using this and the the equality $\lim_{n\to\infty} nu_{n+1} =\lim_{n\to\infty} \frac{n}{n+1} (n+1) u_{n+1} =0$ you get $$ \lim_{n\to\infty} s_{n+1} =\lim_{n\to \infty }\frac{s_1 +2s_2 +...+ns_n}{n}.$$

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  • $\begingroup$ But how can we see that $$\lim_{n\to\infty}\frac {u_2+2u_3+\dots+nu_{n+1}}{n}=0$$ $\endgroup$ – superAnnoyingUser Jun 25 '14 at 11:04
  • $\begingroup$ Because if $\lim_{n\to\infty} a_n$ exists then $$\lim_{n\to\infty} \frac{a_1 +a_2 +...+a_n}{n}=\lim_{n\to\infty} a_n .$$ $\endgroup$ – user110661 Jun 25 '14 at 11:11

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