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How do I determine the value of the following alternating sum (converges by Leibniz):

$$ \sum\limits_{n = 0}^{\infty}\sum\limits_{k = 0}^{2n}{2n\choose k}(-1)^k (p)^{k+1},\quad p \in (0, 1) $$

I don't have any idea how one can tackle such a sum as I just started learning about them.

Jacob
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    Hint: replace $(-1)^k$ by $\pm(-1)^{n-k}$ and make $(1-p)^n$ binomial expansion appear. Are you sure it is $2n$ ? – zwim Mar 30 '21 at 15:38
  • @zwim it's funny that you propose that because that's exactly what the some looked like at first, and then I transformed it in the above while hoping this would make it easier – Jacob Mar 30 '21 at 15:40
  • ah, I see, I made some mistake at first, thanks! – Jacob Mar 30 '21 at 15:41
  • Then set $r=1-p$ and $\sum r^n$ is the sum of a geometric series. – zwim Mar 30 '21 at 15:42
  • @zwim yes, that's what I did! – Jacob Mar 30 '21 at 15:46

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$$ S= \sum\limits_{n = 0}^{\infty}\sum\limits_{k = 0}^{2n}{2n\choose k}(-1)^k (p)^{k+1}$$ $$ \sum\limits_{k = 0}^{2n}{2n\choose k}(-1)^k (p)^{k+1}= p~(1-p)^{2n} $$ $$ S = \sum\limits_{n = 0}^{\infty} p~(1-p)^{2n} = p \cdot \sum\limits_{n = 0}^{\infty} (1-p)^{2n} $$ $$p \in (0, 1) \implies 1-p \in (0, 1) $$ Now, all that's left is the sum of an infinite geometric progression with common ratio $ (1-p)^2 \in (0, 1)$

Ankit Saha
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