# If $a_i\geq 0,$ prove $\sum\limits_{n=1}^\infty\frac{a_1+a_2+\cdots+a_n}{n}$diverges.

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 $a_i\neq 0$.

• By why you say "we all know" (??), I assume your sequence is non-negative? – Timbuc Apr 5 '15 at 15:37
• There are some additional hypothesis missing – Daniel Apr 5 '15 at 15:39
• The $n$-th partial sum is clearly greater than $\sum_{i=1}^n a_1/i = a_1\sum_{i=1}^n 1/i$. – Tad Apr 5 '15 at 15:43
• @Timbuc Yes, it's non-negative. – Ferry Tau Apr 5 '15 at 15:43
• @SpamIAm$a_n\neq0,$Sorry, I forget it. – Ferry Tau Apr 5 '15 at 15:45

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.
• What is $\;a_{n_0}\;$ and how come that mean is greater than $\;a_{n_0}\cdot\frac1n\;$ ?? – Timbuc Apr 5 '15 at 15:45
• Why is $a_i \ge a_{n_0}$? – Keba Apr 5 '15 at 15:46
• Just discard all terms except $a_{n_0}$, and since they are positive... – Siminore Apr 5 '15 at 15:47