if $\sum a_n$ is a divergent series of positive terms then so is $\sum \sqrt{a_n}$

Question

Here's my attempt:

Suppose $(a_n)$ isn't bounded. Then $(\sqrt{a_n})$ won't be bounded either. Hence, $\sum \sqrt{a_n}$ is divergent. Now, if $(a_n)$ is bounded, then for all $n$, $a_n < M$ for some $M>0$. Then $\frac{a_n}{M} < \frac{\sqrt{a_n}}{\sqrt{M}}$ for all $n$. Since, $\sum a_n$ diverges, so must $\sum \sqrt{a_n}$ by the comparison test.

Is this a valid proof? Alternative proofs are welcome.

Answer 1

We assume series with positive terms, so convergence is also absolute convergence.

$\sum b_n$ converges $\implies b_n\to 0$ and in particular $b_n<1$ for $n\gg 1$ large enough.

Multiplying by $b_n$ we get ${b_n}^2<b_n$ thus $\sum {b_n}^2$ converges too.

By contraposition $\sum {b_n}^2$ diverges $\implies \sum b_n$ diverges.

Apply to $b_n=\sqrt{a_n}$.

Answer 2

Use the comparison test and observe that: $\sqrt{a_n}= \dfrac{a_n}{\sqrt{a_n}}\ge \dfrac{a_n}{\sqrt{M}}$ where $M>0$ is the upper bound of $a_n$ in the event $a_n$ is bounded. The case unbounded is mentioned in your analysis already.

Answer 3

Let's define $b_k = \sum_{n = 0}^{k}a_n$. Then $b_k\to\infty$ and $\sqrt{b_k}\to\infty$. We have that $\sqrt{x+y} < \sqrt{x}+\sqrt{y}$. So: $$\sqrt{b_k} = \sqrt{\sum_{n = 0}^{k}a_n} < \sum_{n = 0}^{k}\sqrt{a_n} \implies$$ $$\sum_{n = 0}^{k}\sqrt{a_n}\to\infty$$