Convergence and divergence of three series

Question

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.

Answer 1

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

$$ \sum_{m = \mathbf{1}}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(n + m)^2} = \sum_{n = 1}^{\infty} \frac{1}{n} - \sum_{n = 1}^{\infty} \frac{1}{n^2}.$$