# Convergence, absolute convergence, divergence of series

Let the series $\sum_{n=0}^\infty{\frac{a_n}{3^n}}$ be convergent, but the series $\sum_{n=0}^\infty{\frac{{(-1)^n}{a_n}}{3^n}}$ be divergent.
Show whether:

a)$\sum_{n=0}^\infty{\frac{a_n}{3^n}}$ is absolutely convergent or conditionally convergent

b)$\sum_{n=0}^\infty{\frac{a_n}{2^n}}$ is absolutely/conditionally convergent or divergent

c)$\sum_{n=0}^\infty{\frac{a_n}{4^n}}$ is absolutely/conditionally convergent or divergent

d)Find the convergence radius of $\sum_{n=0}^\infty{\frac{n+2}{n+3}a_nx^n}$

Okay, so far I have solved a), which I found quite easy to do, but I seem to get something wrong at b/c/d could someone please explain how they should be solved as it seems I have blocked completely...

• well I tried the root test, but $a_n$ confuses me a lot Jun 20, 2017 at 14:26

HINT:

We have

$$\limsup_{n\to \infty}\sqrt[n]{\left|\frac{a_n}{3^n}\right|}=1$$

Hence, we must have

$$\limsup_{n\to \infty}\sqrt[n]{\left|a_n\right|}=3$$

• so $|a_n|<3^n$ ? Jun 20, 2017 at 14:46
• Actually, $|a_n|\le 3^n$. Now, can you use this alone to ascertain all of the answers? Not quite. But you can use $\limsup_n \sqrt[n]{|a_n|}=3$ to do so. Jun 20, 2017 at 14:53
• so $\lim_{n\to\infty}{\sqrt[n]{\frac{|a_n|}{3^n}}}=\lim_{n\to\infty}{\frac{\sqrt[n]{|a_n|}}{3}}=\frac{sth}{3}$ ? Jun 20, 2017 at 14:56
• What do you mean by $sth$? Jun 20, 2017 at 14:57
• $sth=\lim sup_{n\to\infty}{\sqrt[n]{|a_n|}}$ i think? Jun 20, 2017 at 15:00

Hint: The series $\sum \frac {a_n}{3^n}$ must cause the ratio/root test to fail. Otherwise, we couldn't have the conditional convergence as in a).

• So what should be my next step? Jun 20, 2017 at 14:31
• Your next step should be to think for a little longer. The ratio test fails only in a very specific case. What does this information tell you about $a_n$? Use this information to apply the ratio test to the other series. Jun 20, 2017 at 14:33