Series absolute convergence, proof I had this problem on my last exam in calculus, I still don't know, how to solve this, so I would appreciate your help.
Let $a_{ij} \in \mathbb{R}$ for $i,j \in \mathbb{N}$. Suppose, that for every $j$ series $S_j = \sum_{i=1}^{\infty}{a_{ij}}$ is absolutely convergent and, for every $i$ exists finite limit (limit $\not= \pm\infty$) $c_i=\lim_{j\to\infty}{a_{ij}}$.
(a) Prove, that if exists such absolutely convergent series $\sum_{i=1}^{\infty}{b_i}$ that $|a_{ij}| \leq |b_i|$ for every $i,j$ then $S_j \to S = \sum_{i=1}^{\infty}{c_i}$, $j \to \infty$
(b) Is argument in (a) true without assumption that series $\sum{}b_i$  with properties given in (a) exists?
 A: First let us look at part (b) to extract some ideas about proving part (a). 
Consider the double sequence $a_{ij} = \delta_{ij}$, that is, $a_{ij} = 1$ if $i = j$ and $a_{ij} = 0$ otherwise. Then $S_j = 1$ for any $j$. On the other hand, $c_i = 0$ for any $i$. So $S = \sum c_i = 0 \neq \lim S_j$. So without the dominating sequence $b_i$ the statement is false.  
(Like Pete says, this is another reflection of something in the continuous case, namely the principle of weak convergence and also Fatou's lemma.)
Now we prove part (a). Fix $\epsilon > 0$. Let $N$ be chosen large enough such that $\sum_{i = N}^\infty |b_i| < \epsilon/4 $, this $N$ exists since $(b_i)$ is absolutely convergent. Consider the partial sums $S_{j,N} = \sum_{i = 1}^{N-1} a_{ij}$. We can then pick $M$ sufficiently large such that for all $j > M$ and  $i < N$, $|a_{ij} - c_i| < \epsilon / (4N)$. 
So in particular, $|S_{j,N} - S_N| < \epsilon / 4$ for $j > M$ where $S_N = \sum_{i = 1}^{N-1} c_i$. 
Now, by assumption of convergence, and the domination of $(b_i)$, you have $|c_i| < |b_i|$. So in particular 
$$ |S - S_N| = | \sum_{N}^{\infty} c_i | \leq \sum_{N}^\infty |b_i| \leq \epsilon/4 $$
We also have 
$$ |S_N - S_{j,N} | < \epsilon / 4, \quad j > M $$
(where $M$ implicitly depends on $N$, and hence on $\epsilon$)  and 
$$ |S_j - S_{j,N} | < \epsilon / 4 $$
So we have that, applying the triangle inequality, that for every fixed $\epsilon > 0$, we can pick $M$ sufficiently large such that
$$ |S_j - S| < \epsilon \qquad \forall j > M$$
Q.E.D.
