# Writing the product $\sum\limits_{r=0}^\infty \frac{z^r}{r!} \sum\limits_{s=0}^\infty \frac{z^{-s}}{s!}$ as a power series in $z$

My lecturer states that the product

$$\sum_{r=0}^\infty \frac{z^r}{r!} \sum_{s=0}^\infty \frac{z^{-s}}{s!}$$

can be written as (with $n = r-s$)

$$\sum_{n=0}^\infty z^n\sum_{r=n}^\infty \frac{1}{r!(r-n)!} + \sum_{n=-\infty}^{-1} z^n \sum_{r=n}^\infty \frac{1}{r!(r-n)!}$$

but I just don't understand how this has happened. Can someone do the algebra step by step please?

This is my attempt:

Set $n=r-s$ then we have $s = r-n$ and

$$\sum_{r=0}^\infty \frac{z^r}{r!} \sum_{r-n=0}^\infty \frac{z^{-r+n}}{(r-n)!} = \sum_{r=0}^\infty \sum_{n=r}^\infty \frac{z^{n}}{r!(r-n)!}$$

• This is simply regrouping every contribution $z^{r-s}$ such that $r-s=n$, for each $n$ in $\mathbb Z$. – Did May 17 '16 at 8:39
• @Did I have edited the question with my attempt. Why has his outer sum got $n$ and inner sum got $r$? – user2850514 May 17 '16 at 8:57
• Right, so the change of variables in the indices is from $(r,s)$ with $r\geqslant0$, $s\geqslant0$ to $(r,n)$ with $n=r-s$, that is, to $(r,n)$ with $r\geqslant0$, $n\leqslant r$, or, equivalently, to $(r,n)$ with $n$ in $\mathbb Z$, $r\geqslant\max(n,0)$. Thus, the double sum is $$\sum_{n=0}^\infty\sum_{r=n}^\infty \frac{z^n}{r!(r-n)!}+\sum_{n=-\infty}^{-1}\sum_{r=0}^\infty \frac{z^n}{r!(r-n)!}.$$ Note the mistake in the claim in your post. – Did May 17 '16 at 10:54

Solution in the form requested:

\begin{align} \sum_{r=0}^\infty\frac{z^r}{r!}\sum_{s=0}^\infty\frac{z^{-s}}{s!} &=\sum_{r=0}^\infty\sum_{s=0}^\infty\frac{z^{r-s}}{r!s!}\\ &=\sum_{r=0}^\infty\sum_{n=r-\infty}^r\frac{z^n}{r!(r-n)!} &&\text{putting n=r-s}\\ &=\sum_{r=0}^\infty\sum_{n=-\infty}^r\frac{z^n}{r!(r-n)!}\\ &=\sum_{r=0}^\infty\sum_{n=-\infty}^{-1}\frac{z^n}{r!(r-n)!}+\sum_{r=0}^\infty\sum_{n=0}^r\frac{z^n}{r!(r-n)!}&&\text{splitting ranges}\\ &=\sum_{r=0}^\infty\sum_{n=-\infty}^{-1}\frac{z^n}{r!(r-n)!}+\sum_{n=0}^\infty\sum_{r=n}^\infty\frac{z^n}{r!(r-n)!} &&\text{swapping ordering order of summation}\\ &=\sum_{n=-\infty}^{-1}\sum_{r=0}^\infty\frac{z^n}{r!(r-n)!}+\sum_{n=0}^\infty\sum_{r=n}^\infty\frac{z^n}{r!(r-n)!}\\ &=\sum_{n=-\infty}^{-1}z^n\sum_{r=0}^\infty\frac{1}{r!(r-n)!}+\sum_{n=0}^\infty z^n\sum_{r=n}^\infty \frac{1}{r!(r-n)!}\qquad\blacksquare \end{align}

Solution posted earlier:

The Cauchy product of two infinte series is given by: $$\sum_{r=0}^\infty f(r) \sum_{s=0}^\infty g(s)= \sum_{k=0}^\infty \sum_{l=0}^k f(l)g(k-l)\\$$

Hence \begin{align} \sum_{r=0}^\infty\overbrace{\frac{z^r}{r!}}^{f(r)}\sum_{s=0}^\infty \overbrace{\frac{z^{-s}}{s!}}^{g(s)}&= \sum_{k=0}^\infty\sum_{l=0}^k \frac{z^l}{l!}\frac{z^{-(k-l)}}{(k-l)!}\\ &=\sum_{k=0}^\infty \frac{z^{-k}}{k!}\sum_{l=0}^kz^{2l}\frac{k!}{l!(k-l)!}\\ &=\sum_{k=0}^\infty \frac{z^{-k}}{k!}\sum_{l=0}^kz^{2l}\binom kl\\ &=\sum_{k=0}^\infty \frac{z^{-k}}{k!}(1+z^2)^k\\ &=\sum_{k=0}^\infty \frac{(z+\frac1z)^k}{k!}\\ &=e^{z+\frac 1z}\qquad\blacksquare \end{align}

In this particular case, it would have been more straightforward to evaluate the summations individually and then multiply them i.e. $$\sum_{r=0}^\infty \frac{z^r}{r!}\sum_{s=0}^\infty \frac{z^{-s}}{s!} =\sum_{r=0}^\infty \frac{z^r}{r!}\sum_{s=0}^\infty \frac{(z^{-1})^s}{s!} =e^z \cdot e^{\frac 1z}=e^{z+\frac 1z}$$

• Hi. You haven't answered my question.. I asked for a specific form. Not the closed form of the series. I started from the function $e^{z+1/z}$ and have been asked to find the general $c_n$ in its Laurent expansion. – user2850514 May 17 '16 at 10:06
• Edited to address your specific request. – hypergeometric May 18 '16 at 8:00