How does a sum of products equal the product of sums here?

Question

Question

In the attached image, they separated the summation in the second step. Is that possible?

I think no because

$$\sum_{i,j=1}^n a_n b_n ≠\sum_{i=1}^n a_n \sum_{j=1}^n b_n $$ for eg: $(1.1+2.2+3.3)≠(1+2+3)(1+2+3)$

Further they also separated the limit. Is that possible?

I think for the limit it is possible. But for the summation case it is not possible. Please explain.

Answer 1

It is possible. In the first step, the example shows $$\sum_{l,k=0}^n \frac{k^2x^ky^l}{l!}=\sum_{l=0}^n \left(\frac{y^l}{l!}\sum_{k=0}^n k^2x^k\right)=\left(\sum_{l=0}^n \frac{y^l}{l!}\right)\left(\sum_{k=0}^n k^2x^k\right)$$ (Think about this: on the LHS each combination of $(l,k)$ is considered, and on the RHS each combination of $(l,k)$ is also included too.)

Remember, $l,k$ are independent.

In the second step, the limit can be separated since $\lim_{a\to a_0,b\to b_0} a\cdot b=\lim_{a\to a_0} a\cdot\lim_{b\to b_0} b$

Answer 2

The first line summation is actually a double summation in the sense that \begin{align*} \sum_{k,\ell=0}^{n}\left(\frac{k^{2}}{x^{k}}\cdot\frac{y^{\ell}}{\ell!}\right)=\sum_{\ell=0}^{n}\sum_{k=0}^{n}\frac{k^{2}}{x^{k}}\cdot\frac{y^{\ell}}{\ell!}=\sum_{\ell=0}^{n}\frac{y^{\ell}}{\ell!}\sum_{k=0}^{n}\frac{k^{2}}{x^{k}}. \end{align*}