Checking on some convergent series I need some verification on the following 2 problems I attemped: 

I have to show that the following series 
  is convergent: $$1-\frac{1}{3 \cdot 4}+\frac{1}{ 5 \cdot 4^2 }-\frac{1}{7 \cdot 4^3}+ \ldots$$ .

My Attempt: I notice that the general term is given by $$\,\,a_n=(-1)^{n}{1 \over {(2n+1)4^n}} \,\,\text{by ignoring the first term of the given series.}$$ Using the fact that An absolutely convergent series is convergent, $$\sum_{1}^{\infty}|a_n|=\sum_{1}^{\infty} {1 \over {(2n+1)4^n}}\le \sum_{1}^{\infty} {1 \over 4^n}=\sum_{1}^{\infty}{1 \over {2^{2n}}}$$ which is clearly convergent by comparing it with the p-series with $p >1$.

I have to show that the following series 
  is convergent:$$1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+ \ldots $$

My Attempt:$$1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+ \ldots \le 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+ \frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\ldots $$. Now, using the fact that $n! >n^2$ for $n \ge 4$ and the fact that omitting first few terms of the series does not affect the characteristics of the series ,we see that  $$\frac{1}{4!}-\frac{1}{6!}+\frac{1}{8!}+ \ldots \le \frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+ \frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}+\ldots =\sum_{4}^{\infty}{1 \over n!} <\sum_{4}^{\infty}{1 \over n^2}$$ and it is clearly convergent by comparing it with the p-series with $p >1$.
Now,I am stuck on the third one. 

I have to show that the following series 
  is convergent:$$\frac{\log 2}{2^2}-\frac{\log 3}{3^2}+\frac{\log 4}{4^2}- \ldots $$ 

I see that $$\frac{\log 2}{2^2}-\frac{\log 3}{3^2}+\frac{\log 4}{4^2}- \ldots \le  \sum_{2}^{\infty} {{\log n} \over {n^2}}= ?? $$
Thanks and regards to all.
 A: The basic result that solves all of these
is this:
If $(a_n)_{n=0}^{\infty}$ is a series of reals
such that
$\forall n\  a_n > a_{n+1} > 0$
and
$\lim_{n \to \infty} a_n = 0$,
then
$\sum_{n=0}^{\infty} (-1)^n a_n$
converges.
This can be proven using the
Cauchy criterion
by showing that,
for any $\epsilon > 0$,
there is an $N(\epsilon)$
such that
if $N(\epsilon) < n < m$
then
$|\sum_{k=n}^m (-1)^k a_k|
< \epsilon$.
This needs both that
$a_n > a_{n+1}$
and
$\lim_{n \to \infty} a_n = 0$.
A: Another approach for the last series is the alternating series test which proves the convergence since

i) $ a_{n+1} < a_n, \quad \forall n\geq 2,  $
ii) $ a_n\longrightarrow_{n\to \infty} 0.  $

Note: You can prove the $a_n$ is decreasing by using the first derivative test for the function

$$ f(x)=\frac{\ln x}{x^2}. $$

A: Hint: For all sufficiently large $n$ (in fact, $n \ge 1$ suffices for this), we have $\ln{n} \le \sqrt{n}$; thus
$$\sum\limits_{n = 2}^{\infty} \frac{\ln n}{n^2} \le \sum\limits_{n = 2}^{\infty} \frac{1}{n^{3/2}}$$
which is a $p$-series.
