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
Best Answer
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.