Prove that $\sum_{1}^{\infty} a_{n} n^{r} \lt \infty \rightarrow \sum_{1}^{\infty} a_{n} \lt \infty \; , r\gt 1$

Question

Let be $\{a_{n}\}$ a sequence of real number / $0\lt a_{n} \le 1 \; \forall n \in \Bbb N $. Prove that $\sum_{1}^{\infty} a_{n} n^{r} \lt \infty \rightarrow \sum_{1}^{\infty} a_{n} \lt \infty \; , r\gt 1$

So, what I did:

So, by the Limit comparison test, which states that if $\lim_{x\to \infty} \frac{a_{n}}{b_{n}} = 0 \; \text{then} \; \text{If} \; \sum_{1}^{\infty} b_{n} \lt \infty \rightarrow \sum_{1}^{\infty} a_{n} \lt \infty$

Then, $lim_{x\to \infty} \frac{a_{n}}{a_{n} n^{r}} = lim_{x\to \infty} \frac{1}{a_{n} n^{r}} = 0$

Because we know that $\sum_{1}^{\infty} a_{n} n^{r} \lt \infty \rightarrow \sum_{1}^{\infty} a_{n} \lt \infty$

So that would be it. I don´t know if it's done correctly. I don't know why the hypothesis that $0\lt a_{n} \le 1$. Thanks in advance.

Answer 1

$0 \leq a_n \leq a_nn^{r}$. Use comparison test.

Your proof is also OK but this is simpler. $a_n \leq 1$ is not required.