Showing $\sum_{k=0}^{\infty} \sum_{i+j=k}a_i b_j = \sum_{k=0}^{\infty} a_k \sum_{k=0}^{\infty} b_k$ I want to show $\sum_{k=0}^{\infty} \sum_{i+j=k}a_i b_j = \sum_{k=0}^{\infty} a_k \sum_{k=0}^{\infty} b_k$ rigoursly.
How I can prove rigorously? (Of course suppose $\sum_{k=0}^{\infty}a_k < \infty$, $\sum_{k=0}^{\infty}b_k < \infty$ condition is required)
 A: The weakest condition under which this holds is with at least one of the series $\sum a_k$ and $\sum b_k$ converging absolutely.
Suppose that $\sum_{k=0}^\infty a_k$ is absolutely convergent and $\sum_{k=0}^\infty b_k$ is convergent, so that there exist bounds $A,B > 0$ such that for all $n \in \mathbb{N}$,
$$\tag{1}\sum_{k=0}^n |a_k| < A, \quad \left|\sum_{k=0}^n b_k\right| < B$$
Define the partial sums $A_n = \sum_{k=0}^n a_k$, $B_n = \sum_{k=0}^n b_k$, and note that
$$C_n = \sum_{k=0}^n\sum_{i+j=k}a_ib_j = \sum_{k=0}^n\sum_{j=0}^ka_jb_{k-j} $$
The objective is to prove that $\lim_{n \to \infty}C_n = \lim_{n \to \infty}A_nB_n = \sum_{k=0}^\infty a_k \sum_{k=0}^\infty b_k. $
We have (looking at a table with entries $a_kb_j$ in row $k$ and column $j$ will make this clear)
$$\tag{2}|C_{2n}- A_nB_n| = \left|\sum_{k=0}^{n-1}a_k\sum_{j=n+1}^{2n-k}b_j + \sum_{k=n+1}^{2n}a_k\sum_{j=0}^{2n-k}b_j\right| \\ \leqslant \sum_{k=0}^{n-1}|a_k|\left|\sum_{j=n+1}^{2n-k}b_j\right| + \sum_{k=n+1}^{2n}|a_k|\left|\sum_{j=0}^{2n-k}b_j\right|$$
By the Cauchy criterion, for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n \geqslant N$ and all $m \geqslant 0$, we have
$$\tag{3}\sum_{k=n+1}^{n+m} |a_k| < \frac{\epsilon}{A+B}, \quad \left|\sum_{j=n+1}^{n+m} b_j\right| < \frac{\epsilon}{A+B}$$
Applying (3) with $m = n-k$ and the bounds in (1) to (2), we get for all $n \geqslant N$
$$|C_{2n}- A_nB_n|\leqslant A \cdot\frac{\epsilon}{A+B} + \frac{\epsilon}{A+B} \cdot B= \epsilon$$
Since $|C_{2n} - \lim_{n \to \infty}A_nB_n| \leqslant |C_{2n} - A_nB_n|+ |A_nB_n - \lim_{n \to \infty}A_nB_n|$ this implies that the subsequence $C_{2n}$ converges to $\lim_{n \to \infty} A_nB_n$. By a similar argument we can show that the subseqence $C_{2n+1}$ converges to $\lim_{n \to \infty} A_{n+1}B_n$.
Q.E.D

The result does not hold if both series $\sum a_k$ and $\sum b_k$ are conditionally convergent.  The standard counterexample is
$$a_k = b_k = \frac{(-1)^{k+1}}{\sqrt{k+1}}$$
A: Let $P(x) = \sum_{i=0}^\infty a_i x^i,$ and let $Q(x) = \sum_{j=0}^\infty b_j x^j.$
Then, your limit just says that
$P(1) Q(1) = (PQ)(1).$ Notice that for this to make sense, both series have to have radius of convergence $1$ (and as pointed out in the other answer, one of them has to converge absolutely at $1.$).
