2
$\begingroup$

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 $\alpha \in \mathbb{R}$.

$Proof.$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – mengdie1982 Nov 12 '19 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 '19 at 15:24
  • $\begingroup$ Moreover,your inequality does not always hold, since we are not given that $\alpha>0$. $\endgroup$ – mengdie1982 Nov 12 '19 at 15:27
  • $\begingroup$ But how you know $(S_n)^\alpha>a_1^{\alpha}$? $\endgroup$ – mengdie1982 Nov 12 '19 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 '19 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.