0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Have you tried any other convergence tests? $\endgroup$ – Andrew Li Feb 20 '18 at 4:47
  • $\begingroup$ 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. $\endgroup$ – RAHUl JHa Feb 20 '18 at 4:51
  • $\begingroup$ but have you actually proved it via tests? $\endgroup$ – Andrew Li Feb 20 '18 at 4:52
  • $\begingroup$ 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 $\endgroup$ – RAHUl JHa Feb 20 '18 at 4:58
  • $\begingroup$ Are you taking a calculus course with sequences & series? They should cover convergence tests for these problems IIRC. $\endgroup$ – Andrew Li Feb 20 '18 at 5:00
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $a_n = \dfrac{1}{n^2}$.

$\endgroup$
  • $\begingroup$ You can't assume $a_n>0$, as the OP didn't mention such an assumption. $\endgroup$ – Professor Vector Feb 20 '18 at 4:58
  • $\begingroup$ Thanks DeepSea, can you give me a counter example of the fourth option. $\endgroup$ – RAHUl JHa Feb 20 '18 at 5:01
  • 1
    $\begingroup$ @Professor Vector if you do not assume that $a_n >0$ how is $\sqrt a_n$ in 2) defined? $\endgroup$ – Kavi Rama Murthy Feb 20 '18 at 5:33
  • $\begingroup$ @Kavi Rama Murthy Good question, but you'd have to ask the author of those problems, not me. $\endgroup$ – Professor Vector Feb 20 '18 at 5:35
  • $\begingroup$ there can be series of complex numbers right? $\endgroup$ – jnyan Feb 20 '18 at 7:17
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.