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 $a_i\neq 0$.

  • 3
    $\begingroup$ 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, 2015 at 15:43
  • $\begingroup$ How do you prove $\sum\limits_{n=1}^\infty \sqrt[n]{a_1a_2\cdots a_n}$ and $\sum\limits_{n=1}^\infty\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}$ converge? Thanks. $\endgroup$
    – Hans
    Apr 9, 2015 at 19:19
  • $\begingroup$ @Hans,The provement may be long and Will I type them here? $\endgroup$
    – Lorence
    Apr 10, 2015 at 14:29
  • $\begingroup$ I have opened a question math.stackexchange.com/questions/1228749/…. You may post your answer there. $\endgroup$
    – Hans
    Apr 10, 2015 at 18:54

1 Answer 1


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.

  • $\begingroup$ What is $\;a_{n_0}\;$ and how come that mean is greater than $\;a_{n_0}\cdot\frac1n\;$ ?? $\endgroup$
    – Timbuc
    Apr 5, 2015 at 15:45
  • 1
    $\begingroup$ Why is $a_i \ge a_{n_0}$? $\endgroup$
    – Keba
    Apr 5, 2015 at 15:46
  • 4
    $\begingroup$ Just discard all terms except $a_{n_0}$, and since they are positive... $\endgroup$
    – Siminore
    Apr 5, 2015 at 15:47
  • $\begingroup$ Ah, my fault. I totally misread your statement. $\endgroup$
    – Keba
    Apr 5, 2015 at 15:50
  • 1
    $\begingroup$ Wow, after seeing this answer I feel incredibly stupid. $\endgroup$ Apr 5, 2015 at 15:59

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