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

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.

• Have you tried any other convergence tests? – Andrew Li Feb 20 '18 at 4:47
• I concluded that first, third and fourth are always true. Am I right? I am not sure. I have taken many counter examples and turned out that these options are true in each counter examples. – RAHUl JHa Feb 20 '18 at 4:51
• but have you actually proved it via tests? – Andrew Li Feb 20 '18 at 4:52
• Actually, I am not comfortable in proving these type of problems formally. I do these by taking counter examples. Can you please tell me how, the fourth one is divergent because answer by Deep sea conclude that the fourth one is divergent. Thanks – RAHUl JHa Feb 20 '18 at 4:58
• Are you taking a calculus course with sequences & series? They should cover convergence tests for these problems IIRC. – Andrew Li Feb 20 '18 at 5:00

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 $\displaystyle\sum\dfrac{1}{n\log(n+1)}=\infty$.

Series $a_n = \frac{(-1)^n}{\sqrt{n}}$ can be a counter example for 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 $a_n = \dfrac{1}{n^2}$.

• You can't assume $a_n>0$, as the OP didn't mention such an assumption. – Professor Vector Feb 20 '18 at 4:58
• Thanks DeepSea, can you give me a counter example of the fourth option. – RAHUl JHa Feb 20 '18 at 5:01
• @Professor Vector if you do not assume that $a_n >0$ how is $\sqrt a_n$ in 2) defined? – Kavi Rama Murthy Feb 20 '18 at 5:33
• @Kavi Rama Murthy Good question, but you'd have to ask the author of those problems, not me. – Professor Vector Feb 20 '18 at 5:35
• there can be series of complex numbers right? – jnyan Feb 20 '18 at 7:17

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.