Determine whether the following series converge and whether they even converge absolutely:
- $\displaystyle\sum\limits_{k=2}^\infty\frac{(-1)^k}{(\log(k))^k}$
- $\displaystyle\sum\limits_{k=1}^\infty\left(\frac{1}{k!}-\frac{3}{(k+1)!}\right)$
- $\displaystyle\sum\limits_{k=1}^\infty\frac{2(k!)^2}{(2k)!}$
- $\displaystyle\sum\limits_{k=1}^\infty\frac{3^{2k-1}}{k^2+k}$
- $\displaystyle\sum\limits_{k=1}^\infty\left(\frac{2+(-1)^k}{4}\right)^k$
- $\displaystyle\sum\limits_{k=1}^\infty(-1)^k\frac{k}{(k+1)(k+2)}$
So I have been able to work through half of the series and I would like to know whether my current attempts are correct and then I would like to get some hints on how to solve the other series I intentionally left out by now.
My current solutions
I know that I have to use the alternating series test and show that $1/(\log(k))^k$ is a null sequence, however I do not know whether it suffices to just mention the monotonicity of $\log$.
$\displaystyle\sum\limits_{k=1}^\infty\left(\frac{1}{k!}-\frac{3}{(k+1)!}\right) = \sum\limits_{k=1}^\infty\left(\frac{k!(k-2)}{k!(k+1)!}\right) = \sum\limits_{k=1}^\infty\left(\frac{k-2}{(k+1)!}\right)$. If we apply the ratio test we get: $\large\left|\frac{\frac{k-1}{(k+2)!}}{\frac{k-2}{(k+1)!}}\right|\normalsize=\frac{k+1}{k^2+4k}=\frac{k(1+1/k)}{k(k-4/k)}\longrightarrow 0<1,\text{ hence the series converges absolutely.}$
The ratio test yields $\large\left|\frac{\frac{2((k+1)!)^2}{(2(k+1))!}}{\frac{2(k!)^2}{(2k)!}}\right| \normalsize = \frac{2((k+1)!)^2(2k)!}{(2(k+1))!2(k!)^2}=\frac{(k+1)^2(2k)!}{(2(k+1))!}=\frac{(k+1)^2}{(2k+2)(2k+1)}=\frac{k^2+2k+1}{4k^2+6k+2}=\frac{k^2(1+2/k+1/k^2)}{k^2(4+6/k+2/k^2)}\longrightarrow \frac{1}{4}<1,\text{ hence the series converges absolutely.}$
The ratio test yields $\large\left|\frac{\frac{3^{2(k+1)-1}}{(k+1)^2+(k+1)}}{\frac{3^{2k-1}}{k^2+k}}\right|\normalsize = \frac{3^{2k+1}(k^2+k)}{3^{2k-1}((k+1)^2+(k+1))} = \frac{9k^2+9k}{k^2+3k+2}=\frac{k^2(9+9/k)}{k^2(1+3/k+2/k^2)}\longrightarrow 9>1,\text{ hence the series diverges.}$
I have no idea at all... a hint might help me out.
Using the alternating series test, we have to show that $\frac{k}{(k+1)(k+2)}$ is a null sequence. This is easily done by $\frac{k}{(k+1)(k+2)}=\frac{1}{k+3+2/k}\longrightarrow 0.$ The series does not converge absolutely based on the comparison test where $\sum\limits_{k=1}^\infty \frac{1}{k+3+2/k}\geq \sum\limits_{k=1}^\infty \frac{1}{k}.$
Thanks for your time and reviewing my attempts.