Convergence and divergence of three series Which of the following is true.
A)$ \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n}$ does not converge. 
B) $ \sum_{n=1}^{\infty} \dfrac{1}{n}$ converges.
C) $ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{(n+m)^2}$ converges. 
D) $ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{(n+m)^2}$ diverges. 
I know that 1st is true but i am confused with 3rd and 4th means is it possible that series is neither convergent nor divergent. 
Any one explain how it is possible.

Note that
\begin{align*}
\sum_{m = \mathbf{0}}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(n + m)^2} &=  \sum_{n=1}^{\infty} \frac{1}{(n + 0 )^2} + \sum_{n=1}^{\infty} \frac{1}{(n  + 1)^2} + \sum_{n=1}^{\infty} \frac{1}{(n + 2 )^2} + \ldots \\\\ &= \sum_{n=1}^{\infty} \frac{1}{(n )^2} + \sum_{n=2}^{\infty} \frac{1}{(n )^2} + \sum_{n=3}^{\infty} \frac{1}{(n )^2} + \ldots \\ &=  \sum_{n = 1}^{\infty} \frac{n}{n^2}
\end{align*}
so 
$$ \sum_{m = \mathbf{1}}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(n + m)^2} =  \sum_{n = 1}^{\infty} \frac{1}{n} - \sum_{n = 1}^{\infty} \frac{1}{n^2}.$$
