0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ You should recognise $r^{-2l}\sum_{k = 0}^l \binom{l}{k}(-1)^kr^k$ if you look at it for a bit. $\endgroup$ – Daniel Fischer Dec 9 '16 at 14:01
  • $\begingroup$ yes your're right thanks. it's $r^{-2l} (1-r)^l $, right? $\endgroup$ – ghthorpe Dec 9 '16 at 14:16
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Dec 9 '16 at 14:21
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Dec 9 '16 at 14:46
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Dec 9 '16 at 14:46
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.