# double summation binomial coefficient

I have the following sum to evaluate: $\sum_\limits{l,k=0}^{\infty} \binom{l}{k} (-1)^kr^{k-2l}$ .

I feel like I first have to establish absolute convergence for a certain range of values of $|r|$ but I'm not sure how to do that.

Next I think I can then use a double summation like so: $\sum_\limits{l=0}^{\infty}\sum_\limits{k=0}^{\infty} \binom{l}{k} (-1)^kr^{k-2l} = \sum_\limits{l=0}^{\infty}\sum_\limits{k=0}^{l} \binom{l}{k} (-1)^kr^{k-2l}$ but I'm not sure how to evaluate the sum from that point on.

• You should recognise $r^{-2l}\sum_{k = 0}^l \binom{l}{k}(-1)^kr^k$ if you look at it for a bit. – Daniel Fischer Dec 9 '16 at 14:01
• yes your're right thanks. it's $r^{-2l} (1-r)^l$, right? – ghthorpe Dec 9 '16 at 14:16
• Right. And we're left with a geometric series. And the task to find out when the manipulations are justified. To find the range of $\lvert r\rvert$ where we have absolute convergence is rather easy. If you need to check whether everything is legitimate for some values outside that range, that can get fussy. – Daniel Fischer Dec 9 '16 at 14:21
• In general, absolute convergence is a sufficient, but not a necessary condition to justify a particular rearrangement. Now here we have $\sum_{l,k = 0}^\infty$, and unless there are special conventions in place for a double-indexed sum, indeed absolute convergence is necessary for the sum to be well-defined [actually, the special form of the terms ensures that all terms are positive when $r < 0$, so in that case the sum is well-defined even if it diverges to $+\infty$. Meh.]. That last remark however points to how to prove things. If all terms are nonnegative, you can rearrange as you please. – Daniel Fischer Dec 9 '16 at 14:46
• The value of the sum, whether finite or $+\infty$ doesn't change under arbitrary rearrangements. So just drop the $(-1)^k$, put absolute value bars around $r$, and rearrange until you get a geometric series, then you have the condition that its ratio must be smaller than $1$ for convergence. – Daniel Fischer Dec 9 '16 at 14:46

Considering $$S=\sum_{l=0}^{\infty}\sum_{k=0}^{\infty} \binom{l}{k} (-1)^kr^{k-2l} =\sum_{l=0}^{\infty}r^{-2l}\left(\sum_{k=0}^{\infty} \binom{l}{k} (-r)^k \right)=\sum_{l=0}^{\infty}{r^{-2l}}{(1-r)^l}$$ $$S=\sum_{l=0}^{\infty}\left(\frac{1-r}{r^2}\right)^l=\frac{r^2}{r^2+r-1}$$