Showing divergence of the series. I am having hard time trying to show that the following series is divergent. Can someone help me please?
$$\sum \frac{(-1)^n}{\log n} b_n $$
where $b_n=\frac{1}{\log n} $  if n is even
and   $b_n=\frac{1}{2^n}$ if n is odd.
This would be a great help!
 A: The Idea: The terms with $n$ odd are negative. But they don't help much, since there is an absolute bound $b$ such that the absolute value of the sum of any finite number of them is $\lt b$. This is because of the rapid decay of $\frac{1}{2^n}$. 
But $\sum_1^m \frac{1}{\log^2(2k)}$ grows without bound, so can be made arbitrarily large, say larger than $C$ for any given $C$. For it is not hard to show by comparison that the series $\sum \frac{1}{\log^2(2k)}$ diverges.  
Thus the partial sum of our series can be made larger than $C-b$ for any $C$.
A: Your series is: 
$$
\frac{1}{\log{2}}\frac{1}{\log{2}} - \frac{1}{\log{3}}\frac{1}{2^3} + \frac{1}{\log{4}}\frac{1}{\log{4}}-\frac{1}{\log{5}}\frac{1}{2^5}+-\ldots 
$$
Which is: 
$$
\sum_{2}^{\infty}\left(\frac{1}{\log{2n}}\right)^2 - \sum_{2}^{\infty} \frac{1}{\log{(2n+1)}2^{2n+1}}
$$
$$
\sum_{2}^{\infty} \frac{1}{\log{(2n+1)}2^{2n+1}}   < \sum_{2}^{\infty} \frac{1}{2^{2n+1}} = \frac{1}{24}
$$
So
$$
\sum_{2}^{\infty}\left(\frac{1}{\log{2n}}\right)^2 - \sum_{2}^{\infty} \frac{1}{\log{(2n+1)}2^{2n+1}} > \left(\sum_{2}^{\infty}\left(\frac{1}{\log{2n}}\right)^2\right) - \frac{1}{24} =\\
\infty - \frac{1}{24} = \infty
$$
