A question about convergent series I need help with the following:
We are given the series $\sum_1^\infty \frac{a_n}{5^n}$ and $\sum_1^\infty \frac{(-1)^na_n}{5^n}$ . We also know that the first series converges while the second one diverges. Now we must answer these questions:
1) is series $\sum_1^\infty \frac{a_n}{5^n}$ a)absolutely convergent or b)conditionally convergent?
2) is the series $\sum_1^\infty \frac{a_n}{4^n}$ a)absolutely convergent, b)conditionally convergent, c) divergent
3) is the series $\sum_1^\infty \frac{a_n}{6^n}$ a)absolutely convergent, b)conditionally convergent, c) divergent
4) What is the radius of convergence of is the series $\sum_1^\infty (n+1)a_nx^n$
1b) seems obvious and I can easily see that 2a) is wrong by the comparison test. But how can I find the exact answer? I tried using the series test, but it doesn't give me information about conditional divergance.
Any help would be appreciated
 A: Re 2., if the sequence $\left(\frac{a_n}{4^n}\right)_{n\geqslant0}$ is bounded, then $|a_n|\leqslant c\cdot4^n$ for some finite $c$, for every $n\geqslant0$, hence $\left|\frac{a_n}{5^n}\right|\leqslant c\cdot\left(\frac45\right)^n$ and $\sum\limits_n\frac{a_n}{5^n}$ is absolutely convergent. This proves that $\left(\frac{a_n}{4^n}\right)_{n\geqslant0}$ is not bounded hence $\sum\limits_n\frac{a_n}{4^n}$ diverges (2.c).
Re 3., $\sum\limits_n\frac{a_n}{5^n}$ converges hence $\left(\frac{a_n}{5^n}\right)_{n\geqslant0}$ is bounded, hence $\left|\frac{a_n}{6^n}\right|\leqslant c\cdot\left(\frac56\right)^n$ for some finite $c$, for every $n\geqslant0$, and $\sum\limits_n\frac{a_n}{6^n}$ converges absolutely (3.a).
Re 4., the radius of convergence of the series $\sum\limits_n(n+1)a_nx^n$ and $\sum\limits_na_nx^n$ are the same (always). The divergence result of 2. holds for every $x\gt\frac15$ instead of $\frac14$ and the absolute convergence result of 3. holds for every $x\lt\frac15$ instead of $\frac16$.
Hence the radius of convergence is $R=\frac15$.
A: Hint: if $$\limsup_{n\to\infty}\left(\frac{|a_n|}{5^n}\right)^{1/n}< 1$$ then the series $\sum_1^\infty \frac{a_n}{5^n}$ would converge absolutely.
