Given $a_n>0$, and $\sum_{n=1}^{\infty}a_n$ converges. Let $$A_m=\frac{1}{m}\sum_{n=1}^{m}a_n\ ,\ B_m=\frac{m}{\sum_{n=1}^{m}a_n^{-1}}$$ representing arithmetic mean and harmonic mean.It's trivial that $\sum_{m=1}^{\infty}A_m$ is not convergent.However, does $\sum_{m=1}^{\infty}B_m$ converge?And if we define $$H_{m}^{s}=\left(\frac{a_1^s+a_2^s+\cdots+a_m^s}{m}\right)^{\frac{1}{s}}$$ where $s\neq0$. Can we find all values of $s$ that makes $\sum_{m=1}^{\infty}H_{m}^{s}$ converge?
Asked
Active
Viewed 182 times
2
-
For large $m$ I think the terms will start to look like $1/m$, so I have a feeling the final sum does not converge. – K.defaoite Aug 30 '21 at 13:54
-
Is it trivial that $\sum_{m=1}^{\infty}A_m$ is not convergent? I don't see it yet, perhaps I am missing something obvious. – Martin R Aug 30 '21 at 14:22
-
1@MartinR As $a_n>0$ we have $A_m >a_1/m$ which diverges. – Severin Schraven Aug 30 '21 at 14:27
-
1@SeverinSchraven: Yes, of course. – Martin R Aug 30 '21 at 14:28
-
Then $\sum_{m=1}^{\infty}H_{m}^{s}$ “trivially” diverges for $s \ge 1$. Convergence for $s \le 0$ (and in particular for $\sum B_m$) is demonstrated here: https://math.stackexchange.com/q/1228749/42969. – Martin R Aug 30 '21 at 15:12
-
The case $0 < s < 1$ is treated here: https://math.stackexchange.com/q/2587408/42969. – Martin R Sep 21 '21 at 16:53