# Proof-Verification： $\sum_{n=1}^{\infty}\frac{a_n}{(S_n)^{\alpha}}$ is convergent.

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 $$\dfrac{L}{2} Therefore, $$(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.

That is correct. You could also argue that $$0 < a_1 \le S_n \le L$$ for all $$n$$ and therefore $$0 < \frac{1}{(S_n)^{\alpha}} \le \max \left( \frac{1}{L^\alpha}, \frac{1}{(a_1)^\alpha} \right) \, .$$
• 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? – mengdie1982 Nov 12 '19 at 15:23
• Moreover,your inequality does not always hold, since we are not given that $\alpha>0$. – mengdie1982 Nov 12 '19 at 15:27
• But how you know $(S_n)^\alpha>a_1^{\alpha}$? – mengdie1982 Nov 12 '19 at 15:29
• @mengdie1982: Yes, you are right. – So there is not much to improve in your proof, a minimal simplification would be $a_1 \le S_n \le L$. – Martin R Nov 12 '19 at 15:30