Verifaction of convergence/divergence exercise I have the following assignment in my textbok:
Series $\sum_{n=0}^{\infty}c_{n}3^n$ is convergent. Based on that can we conclude that the following series coverge:


*

*a) $\sum_{n=0}^{\infty}c_{n}2^n$

*b) $\sum_{n=0}^{\infty}|c_{n}|2^n$

*c) $\sum_{n=0}^{\infty}c_{n}(-3)^n$
a) $c_{n}2^n < c_{n}3^n$
$2^n < 3^n$
$2 < 3$
and since $\sum_{n=0}^{\infty}c_{n}3^n$, $\sum_{n=0}^{\infty}c_{n}2^n$ also converges
b)even though $\sum_{n=0}^{\infty}c_{n}2^n$ converges, it doesn't mean that its absolute values converges and there for I can't say anything about whether this seris converges or diverges
c)even though $\sum_{n=0}^{\infty}c_{n}3^n$ converges, that doesn't mean that its absolute value will converge and there for I can't say anything about whether this seris converges or diverges
Are my conclusions right and mathematical rigorous? 
 A: For 1, your logic is backward, but that's not the main problem--the main problem is that this argument fails for $c_n < 0$.
For 2, you are exactly correct--convergence does not imply absolute convergence. As well, it passes the divergence test ($c_n3^n$ must approach $0$, so $c_n$ must approach $0$), so we cannot conclude either way.
For 3, you are right, but for the wrong reason--this series is equivalent to $$\sum_{n=0}^{\infty}(-1)^nc_n3^n,$$ and since the original series converges, we know that $c_n3^n$ must approach $0$. However, we do not know that $c_n3^n$ is a monotonically decreasing series, so we cannot conclude its convergence.
A: a) $c_n2^n<c_n3^n$ implies $\sum c_n3^n$ converge work only if $c_n>0$. But here you can't say anything.
b) Your argument is correct.
c) This one converge. Indeed, the two series $-\sum c_n 3^n$ and $\sum c_n 3^n$ converge, $$-c_n3^n\leq (-1)^n c_n3^n\leq c_n 3^n$$ or $$c_n3^n\leq (-1)^n c_n3^n\leq -c_n 3^n$$ for all $n$, and because $\lim_{n\to\infty }(-1)^nc_n3^n =0$.
