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If $a_i\geq0,a_n\not\equiv 0,$ prove $\sum\limits_{n=1}^\infty\frac{a_1+a_2+\cdots+a_n}{n}$ diverges.

I have known that if $\sum\limits_{n=1}^\infty a_n$converges, then $\sum\limits_{n=1}^\infty \sqrt[n]{a_1a_2\cdots a_n}$and $\sum\limits_{n=1}^\infty\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}$converges, but how can I prove $\sum\limits_{n=1}^\infty\frac{a_1+a_2+\cdots+a_n}{n}$ diverges?

Sincerely thanks for your help.

Add : Such stupid as I am, we do not need the condition $\sum a_n$ converges.

Strictly, rewrite $\sum\limits_{n=1}^\infty\frac{a_1+a_2+\cdots+a_n}{n}=\sum\limits_{n=k}^\infty\frac{a_1+a_2+\cdots+a_n}{n}\geq a_k\sum\limits_{n=k}^\infty\frac 1n\to \infty$, where $a_k$ is the first $a_i\neq 0$.

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    $\begingroup$ By why you say "we all know" (??), I assume your sequence is non-negative? $\endgroup$ – Timbuc Apr 5 '15 at 15:37
  • $\begingroup$ There are some additional hypothesis missing $\endgroup$ – Daniel Apr 5 '15 at 15:39
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    $\begingroup$ The $n$-th partial sum is clearly greater than $\sum_{i=1}^n a_1/i = a_1\sum_{i=1}^n 1/i$. $\endgroup$ – Tad Apr 5 '15 at 15:43
  • $\begingroup$ @Timbuc Yes, it's non-negative. $\endgroup$ – Ferry Tau Apr 5 '15 at 15:43
  • $\begingroup$ @SpamIAm$a_n\neq0,$Sorry, I forget it. $\endgroup$ – Ferry Tau Apr 5 '15 at 15:45
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Suppose $a_{n_0}\neq 0$, then for all $n\geq n_0$, $$\frac{a_1+\cdots+a_n}{n}\geq a_{n_0}\times\frac{1}{n}$$ hence diverges since the harmonic series diverges.

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  • $\begingroup$ What is $\;a_{n_0}\;$ and how come that mean is greater than $\;a_{n_0}\cdot\frac1n\;$ ?? $\endgroup$ – Timbuc Apr 5 '15 at 15:45
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    $\begingroup$ Why is $a_i \ge a_{n_0}$? $\endgroup$ – Keba Apr 5 '15 at 15:46
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    $\begingroup$ Just discard all terms except $a_{n_0}$, and since they are positive... $\endgroup$ – Siminore Apr 5 '15 at 15:47
  • $\begingroup$ Ah, my fault. I totally misread your statement. $\endgroup$ – Keba Apr 5 '15 at 15:50
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    $\begingroup$ Wow, after seeing this answer I feel incredibly stupid. $\endgroup$ – Kayle of the Creeks Apr 5 '15 at 15:59

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