calculus - Evaluate $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)$

I'd like to Prove that $\sum\limits_{n=0}^{\infty}\sum\limits_{r=0}^{n}\left(\frac{1}{(n-r)!}a^{n-r}\right)\left(\frac{1}{r!}b^{r}\right)=\left(\sum\limits_{n=0}^{\infty}\frac{1}{n!}a^n\right)\left(\sum\limits_{n=0}^{\infty}\frac{1}{n!}b^n\right)$

I do as follow

$\sum\limits_{n=0}^{\infty}\sum\limits_{r=0}^{n}\left(\frac{1}{(n-r)!}a^{n-r}\right)\left(\frac{1}{r!}b^{r}\right)=\sum\limits_{r=0}^{0}\left(\frac{1}{(0-r)!}a^{0-r}\right)\left(\frac{1}{r!}b^{r}\right)+\sum\limits_{r=0}^{1}\left(\frac{1}{(1-r)!}a^{1-r}\right)\left(\frac{1}{r!}b^{r}\right)+\sum\limits_{r=0}^{2}\left(\frac{1}{(2-r)!}a^{2-r}\right)\left(\frac{1}{r!}b^{r}\right)+\cdots$

I couldn't able to get the right hand

Any help will be appreciated! Thanks

Answer

You might find it easier to start from the RHS and show that

$$\left(\sum_{n=0}^{\infty}\frac{1}{n!}a^n\right)\left(\sum_{n=0}^{\infty}\frac{1}{n!}b^n\right) = \sum\limits_{n=0}^{\infty}\sum_{r=0}^{n}\left(\frac{1}{(n-r)!}a^{n-r}\right)\left(\frac{1}{r!}b^{r}\right).$$

Actually, for me this is the only step. It's just how you multiply two series.

Something more interesting would be to see what you can make of

$$\frac{1}{(n-r)!r!}a^{n - r}b^r = \frac{1}{n!} \binom{n}{r} a^{n - r}b^r$$

and the Binomial Theorem.