Which of these series will be always convergent if $\sum a_n$ is convergent.?

Question

Let the series $\sum a_n$ be convergent. Then which of the following will always be convergent? $$\sum \left(a_n\right)^2\tag1$$ $$\sum \sqrt{a_n}\tag2$$ $$\sum \frac{\sqrt{a_n}}{n}\tag3$$ $$\sum \frac{\sqrt{a_n}}{n^{1/4}}\tag4$$

I think we can immediately tell that the second option is not convergent since $\sum \frac{1}{n^2}$ is a counter example. But I am not sure about the remaining three. Any help would be appreciated.

Answer 1

Assume that your series has positive terms: $a_n > 0$ for $n \ge 1$, then $a_n \to 0 \implies a_n < 1$ for $n \ge K \implies a_n^2 < a_n$. Thus $1)$ and $3)$ are always convergent. $3)$ is convergent by AM-GM inequality. $2)$ is divergent for $a_n = \dfrac{1}{n^2}$.

Answer 2

For d), try $a_{n}=\dfrac{1}{n(\log (n+1))^{2}}$, then $\dfrac{\sqrt{a_{n}}}{n^{1/4}}=\dfrac{1}{n^{3/4}\log(n+1)}\geq\dfrac{1}{n\log(n+1)}$ and $\displaystyle\sum\dfrac{1}{n\log(n+1)}=\infty$.

Answer 3

Series $a_n = \frac{(-1)^n}{\sqrt{n}}$ can be a counter example for 1)

Answer 4

assuming positive $a_n$ terms:

First one converges since, after some terms, $(a_n)^2 $ is less than $a_n$

second one doesnt converge always, and your counter example works

third converges using A.M G.M inequality with $a_n$ and $\frac1{n^2}$

fourth doesnt converge always, $\frac1{n^{3/2}}$ will do for counterexample.