Double series with factorials I don't figure out how to compute it. Some help pls?
$$ \sum_{i=1}^{\infty} \sum_{j=i-1}^{\infty} \frac{3^j}{j!}$$
 A: You can change the order of summation thusly:
$$\sum_{i=1}^{\infty} \sum_{j=i-1}^{\infty} \frac{3^j}{j!} = \sum_{j=0}^{\infty} \frac{3^j}{j!} \sum_{i=1}^{j+1} 1 = \sum_{j=0}^{\infty} \frac{3^j}{j!}  (j+1)$$
A: Hint:
$$
\sum_{i=1}^\infty \sum_{j=i-1}^\infty\frac{3^j}{j!}=\sum_{j=0}^\infty \sum_{i=1}^{j+1}\frac{3^j}{j!}
$$
A: Here's something that might satisfy your wish to avoid pictures. It needs a bit of justification, which I didn't provide (i.e. some thought must be put into defining summation and deriving its properties), but I hope it helps.
Define a set $$A=\{(i,j)\in\mathbb N\times\mathbb N_0|\;j\geq i-1\}.$$ This will be the index set over which the summation will be done. For each $i_0\in\mathbb N$ let $$A_{i_0}=\{(i,j)\in A|\;i=i_0\}$$ and for each $j_0\in\mathbb N_0$ let $$A^{j_0}=\{(i,j)\in A|\;j=j_0\}.$$
You may visualize $A_{i_0}$ as the "$i_0$-th vertical slice" of $A$ and $A^{j_0}$ as "$j_0$-th horizontal slice" of $A$.
Note that $$\bigcup_{i=1}^\infty A_i=A=\bigcup_{j=0}^\infty A^{j}\tag{*}$$ which you can prove by checking that these sets contain the same elements. Summation (if defined in the appropriate sense, i.e. Lebesgue integral with respect to the counting measure) is countably additive, which basically means that you will get the same answer if you:


*

*cut up your index set into countably many pieces, sum over each piece, and sum the results, or

*sum over the entire index set in the first place.


Therefore, $(*)$ implies $$\sum_{i=1}^\infty\sum_{(i,j)\in A_i} \frac{3^j}{j!}=\sum_{(i,j)\in A}\frac{3^j}{j!}=\sum_{j=0}^\infty\sum_{(i,j)\in A^j} \frac{3^j}{j!}.\tag{**}$$
Now, note that $$\sum_{(i,j)\in A_i} \frac{3^j}{j!}=\sum_{j=i-1}^\infty \frac{3^j}{j!},$$ since $(i,j)\in A_i$ holds for a fixed $i$ precisely if $j\geq i-1$, i.e. for all $j$ from $i-1$ to infinity, and that $$\sum_{(i,j)\in A^j} \frac{3^j}{j!}=\sum_{i=1}^{j+1} \frac{3^j}{j!},$$ since $(i,j)\in A^j$ holds for a fixed $j$ precisely if $j\geq i-1$, i.e. it holds for all $i$ that are less than or equal to $j+1$. But then $(**)$ means precisely that $$\sum_{i=1}^\infty\sum_{j=i-1}^\infty \frac{3^j}{j!}=\sum_{j=0}^\infty\sum_{i=1}^{j+1} \frac{3^j}{j!},$$ which is the change of the order of summation we wanted.
Comment. The change of the order may be justified also under the usual definition of summation, but I strongly suspect that what I've written is conceptually simpler.
