Suppose $a_n>0(n=1,2,\cdots)$$S_n=\sum\limits_{k=1}^na_n,$ and $\sum\limits_{n=1}^{\infty}a_n $ is convergent. Prove $\sum\limits_{n=1}^{\infty}\dfrac{a_n}{(S_n)^{\alpha}}$ is also convergent for any $\alpha \in \mathbb{R}$.


Denote $\sum\limits_{n =1}^{\infty}a_n=\lim\limits_{n \to \infty}S_n=L.$ Then $ L\geq a_1>0. $ Thus, for a sufficiently large $n$, it holds that $\dfrac{L}{2}<S_n<L.$ Therefore, $(S_n)^{\alpha} $ is always bounded for any $\alpha \in \mathbb{R}$, which implies $\dfrac{1}{(S_n)^{\alpha}} $ is also bounded. Let $M$ be an upper bound of it. Then we obtain $\dfrac{a_n}{(S_n)^{\alpha}}\leq Ma_n$. By the comparison test, the conclusion is followed.


1 Answer 1


That is correct. You could also argue that $0 < a_1 \le S_n \le L$ for all $n$ and therefore $$ 0 < \frac{1}{(S_n)^{\alpha}} \le \max \left( \frac{1}{L^\alpha}, \frac{1}{(a_1)^\alpha} \right) \, . $$

  • $\begingroup$ But if we are not given the convergence of $ \sum_{n=1}^{\infty}a_n$. the series which consists of the right hand of your inequality is convergent? $\endgroup$ Nov 12, 2019 at 15:23
  • $\begingroup$ @mengdie1982: Yes, I had already removed that remark again. (It seems that there was some conflict between your and my edits.) $\endgroup$
    – Martin R
    Nov 12, 2019 at 15:24
  • $\begingroup$ Moreover,your inequality does not always hold, since we are not given that $\alpha>0$. $\endgroup$ Nov 12, 2019 at 15:27
  • $\begingroup$ But how you know $(S_n)^\alpha>a_1^{\alpha}$? $\endgroup$ Nov 12, 2019 at 15:29
  • $\begingroup$ @mengdie1982: Yes, you are right. – So there is not much to improve in your proof, a minimal simplification would be $a_1 \le S_n \le L$. $\endgroup$
    – Martin R
    Nov 12, 2019 at 15:30

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