Proof of convergence with unspecified $a_n$ in sum

Question

The problem is as follows:

Let $a_n \geq 0$. If $$ \sum_{n=1}^{\infty} a_n <\ \infty,$$ show that $$\sum_{n=1}^{\infty} (a_n)^{\frac{n}{n+1}} < \infty.$$

I'm not sure how to approach this. If I use a ratio test, I have an $a_{n+1}$ term that I don't know what to do with. Can I prove that $a_n > a_{n+1}$? If so, can I say that

$$\lim_{n \to \infty} \left\lvert\frac{(a_{n+1})^\frac{n+1}{n+2}}{(a_n)^\frac{n}{n+1}}\right\rvert$$

is < 1, and therefore solve the problem via ratio test?

Thanks in advance.

Answer 1

Define $$M_1 = \{n \in \mathbb{N} \ | \ \sqrt[n+1]{a_n} \leq \frac{1}{2} \} \ \text{and} \ M_2 = \{n \in \mathbb{N} \ | \ \sqrt[n+1]{a_n} > \frac{1}{2} \}$$ For $n \in M_1$: $a_n \leq \frac{1}{2^{n+1}}$ and therefore $(a_n)^{\frac{n}{n+1}} \leq \frac{1}{2^n}$.

For $n \in M_2$: $(a_n)^{\frac{n}{n+1}} = a_n\cdot\frac{1}{\sqrt[n+1]{a_n}} < 2a_n$.

So, $(a_n)^{\frac{n}{n+1}} \leq \max{(\frac{1}{2^n},2a_n)}=:b_n$ for all $n \in \mathbb{N}$. Since $ \sum_{n=1}^{\infty} a_n <\ \infty$, it is $$\sum_{n=1}^{\infty} (a_n)^{\frac{n}{n+1}} \leq \sum_{n=1}^{\infty} b_n <\ \infty$$