Convergence of a series with a parameter For which $a >0$ does series 
$$ \sum_{n=1}^{\infty} \frac{(1+\cos\frac{1}{n})(-1)^{n+[a]}}{(1+\ln(n))^a} $$
converge?
I think I'm doing something wrong here but I can't see where.
Let's test convergence:
$$ \frac{(1+\cos\frac{1}{n})(-1)^{n+[a]}}{(1+\ln(n))^a} \leq \frac{2\cdot (-1)^{n+[a]}}{(1+\ln(n))^a} $$
because $\cos(x) \leq 1$. Since I'm only testing convergence here, and multiplying series by a constant doesn't change its convergence, I'll just divide this $2$ out.
Now I have
$$ \frac{(-1)^{n+[a]}}{(1+\ln(n))^a} $$
$ (-1)^{n+[a] } $ behaves almost identically to $ (-1)^{n} $ ( if $a$ is odd it just becomes $-(-1)^{n}$ which still will be alternating ones, so I can apply Leibniz's test.
Denominator goes to infinity as $n \rightarrow \infty $, so limit will be $0$. Now for monotonicness:
$$ \frac{1}{(1+\ln (n))^a} \geq \frac{1}{(1+\ln(n+1))^a} $$
$$ (1 + \ln(n+1))^a \geq (1 + \ln(n))^a $$
both sides are positive so I can take $a$th root
$$ \ln(n+1) \geq \ln(n) $$
This is true for positive $n$s, so the test is also true for them, so the series will always be convergent, independently of $a$?
 A: You are right, the series converges no matter what $a>0$ is. But your proof is wrong: you cannot apply directly Leibniz's test (alternating series test) here. What you did is majoring the absolute value (your inequality is not true with the $(-1)^n$ still in there...), then apply the test to the upper bound you get: but this is not a legit thing to do, since convergence of the new series does not imply convergence of yours. (See in the comments below for a counter example.)

Another way:
First, note that
$$
\frac{(1+\cos\frac{1}{n})(-1)^{n+[a]}}{(1+\ln(n))^a}
= (-1)^{[a]}\frac{(1+\cos\frac{1}{n})(-1)^{n}}{(1+\ln(n))^a}
$$
so for the rest I'll ignore the factor $(-1)^{[a]}$, that has no influence whatsoever  on the convergence.
Since the series is alternating, a natural idea is to apply the alternating series test, which asserts that if $a_n$ is non-negative and non-increasing, converging to $0$, then the series $\sum_n (-1)^n a_n$ converges. Sadly, the above series above does not satisfy the non-increasing part a priori. So we'll expand it into a first term that does, and a second that we are able to analyze using different tools:
$$
\begin{align}
\frac{1+\cos\frac{1}{n}}{{(1+\ln(n))^a}}
&=
\frac{2+\frac{1}{2n^2}+ o\left(\frac{1}{n^2}\right)}{{(1+\ln(n))^a}} \\
&= 
\frac{2}{{(1+\ln(n))^a}}
+
\frac{1+o(1)}{{2n^2(1+\ln(n))^a}}\\
&= 
\frac{2}{{(1+\ln(n))^a}}
+
\frac{1+o(1)}{2n^2 \ln(n)^a}
\end{align}
$$
so that
$$
\begin{align}
(-1)^n \frac{1+\cos\frac{1}{n}}{{(1+\ln(n))^a}}
&=
\frac{2(-1)^n}{{(1+\ln(n))^a}}
+
(-1)^n\frac{1+o(1)}{2n^2 \ln(n)^a}.
\end{align}
$$
Now, for any $a>0$:


*

*the first term is the general term of a convergent series, since we can now apply the alternating series test to it:  $\sum \frac{2(-1)^n}{{(1+\ln(n))^a}}$ is a convergent series

*the second term is absolutely convergent, e.g. by comparison to $\sum \frac{1}{n^2}$


and therefore our original series converges as the sum of two convergent series, no matter what $a>0$ is.
