Is this sum equality always holds? Suppose I know $\sum_{i=1}^{\infty}a_i$ converges to a value, can I write $$\lim_{n\to\infty}\sum_{i=1}^{n}a_i=\sum_{i=1}^{\infty}a_i$$
Also If I know that $\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{ij}$ converges to a value, can I write $$\lim_{n\to\infty}\sum_{i=1}^{n}\left(\lim_{m\to\infty}\sum_{j=1}^{m}a_{ij}\right)=\lim_{m\to\infty}\sum_{j=1}^{n}\left(\lim_{n\to\infty}\sum_{i=1}^{m}a_{ij}\right)=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{ij}$$
Please someone explain this.
 A: The very definition of $\sum _{i = 1}^{\infty } {a}_{i}$ is
$$\sum _{i = 1}^{\infty } {a}_{i} = {\lim }_{n \rightarrow  \infty } \sum _{i = 1}^{n} {a}_{i}$$
when the limit exist.
For the double series, let
$${S}_{n m} = \sum _{i = 1}^{n} \sum _{j = 1}^{m} {a}_{i j}$$
A sufficient condition for ${\lim }_{n \rightarrow  \infty } \left({\lim }_{m \rightarrow  \infty } {S}_{n m}\right) = {\lim }_{m \rightarrow  \infty } \left({\lim }_{n \rightarrow  \infty } {S}_{n m}\right)$ is
$$\sum _{i = 1}^{n} \sum _{j = 1}^{m} \left|{a}_{i j}\right|  \leqslant  C  <  \infty $$
where $C$ does not depend on $n , m$.
There are weaker conditions such as both limits exist and for all ${\epsilon}  >  0$,
there exist ${N}_{{\epsilon}}$ and ${M}_{{\epsilon}}$ such that
$$\left|{S}_{n+i , m+j}-{S}_{n , m}\right|\le \epsilon$$
for all $n  \geqslant  {N}_{{\epsilon}} , m  \geqslant  {M}_{{\epsilon}} , i  \geqslant  0 , j  \geqslant  0$.
Counterexamples are easy to create because of the identity
$${a}_{i , j} = {S}_{i , j}+{S}_{i-1 , j-1}-{S}_{i , j-1}-{S}_{i-1 , j}$$
with the convention that ${S}_{p , q} = 0$ when $p = 0$ or $q = 0$.
So if we take for example
$${S}_{n , m} = \frac{n-m}{n+m+3}$$
we have a counterexample as ${\lim }_{n \rightarrow  \infty } \left({\lim }_{m \rightarrow  \infty } {S}_{n m}\right) =-1$
and ${\lim }_{m \rightarrow  \infty } \left({\lim }_{n \rightarrow  \infty } {S}_{n m}\right) = 1$.
