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

Question

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 1

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.

Answer 2

Hint

Consider the inner summation and show that $$\sum\limits_{r=0}^{n}\left(\frac{a^{n-r}}{(n-r)!}\right)\left(\frac{b^{r}}{r!}\right)=\frac{(a+b)^n}{n!}$$ Now, the lhs will be very familiar to you and the remaining is simple.

Answer 3

By hint Claude Leibovici,

$ \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_{n=0}^{\infty}\frac{1}{n!}(a+b)^n=e^{a+b}$ on the other hand

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

Please let me know if this makes sense to you