Decide whether the series are absolutely convergent , conditionally convergent, or divergent. Decide  whether the series are  absolutely  convergent, conditionally convergent, or divergent.
$$a) \sum _{n=1}^{\infty} (-1)^n \frac{(\log n)^{\log n}}{n^a} , a>0$$
$$b) \sum_{n=1}^{\infty} \frac {(-1)^{[\log n]}}{n}$$ where  [ ]  is  greatest integer  symbol.
My attempt :  the  easiest  way  to  think  about  this  problem is the Leibniz test both $a)$ and $b)$ will be conditionally convergents.
Is  it correct ?
any  hints/solutions
thanks u
 A: For the second part note that for $n$ between $e^{k}$ and $e ^{k+1}$, $(-1)^{[\log n]}$ has  a constant sign. Compare $\sum_{[e^{k}]}^{[e^{k}+1]} \frac  1  n$ with the integral of $\frac 1 x $ from $[e^{k}]$ to $[e^{k+1}]$ to see that the sum of the terms from $[e^{k}]$ to $[e^{k+1}]$ does not tend to $0$. Hence the series is divergent. 
A: For the first one we have that
$$\frac{(\log n)^{\log n}}{n^a}\to \infty$$
indeed by $x=e^y \to \infty$ as $y\ge e^{2a}$
$$\frac{(\log x)^{\log x}}{x^a}=\frac{y^y}{e^{ay}}\ge\frac{e^{2ay}}{e^{ay}}=e^{ay}\to \infty$$
therefore the given series diverges.
For the second one we can write
$$\sum_{n=1}^{\infty} \frac {(-1)^{[\log n]}}{n}=\sum_{n=1}^{1} \frac {1}{n}-\sum_{n=2}^{2} \frac {1}{n}+\sum_{n=3}^{7} \frac {1}{n}-\sum_{n=8}^{20} \frac {1}{n}+\ldots+(-1)^{k+1}\sum_{n=[e^k]}^{[e^{k+1}-1]} \frac {1}{n}+\ldots =\sum_{k=1}^\infty (-1)^{k+1} a_k$$
that is an alternating sum of terms not convergent to zero and thus it diverges too, indeed as $k\to \infty$ by harmonic series
$$a_k=\sum_{n=[e^k]}^{[e^{k+1}-1]} \frac {1}{n}\sim \log (e^{k+1})-\log(e^k)=k+1-k=1$$
