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Let $p$ in $(0,1)$. How to compute $$\sum_{k=1}^{\infty} 2k{2k\choose k+1} p^{k-1} (1-p)^{k+1}\ ?$$

Personal tries: I've tried to use generating functions, but I can't deal with this $2 k$ in front of the binomial coefficient ${2k\choose k+1}$.

Did
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Mark Tuliy
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  • Your problem statement is not clear. What do you mean by $C_{2k}^{k+1}$? Is it the binomial-coefficient? Even if this is a very often used expression you should still define it because it is not necessarily standard notation. – MrYouMath May 09 '18 at 07:24
  • Also, are you implying $q = 1-p $ or something similar? – Matti P. May 09 '18 at 07:27
  • @MattiP.: I don't see why this would be implied, and in any case, you can easily condense $r:=pq$. –  May 09 '18 at 07:28
  • Context? Personal tries? – Did May 09 '18 at 07:36
  • @Did I've tried to use generating functions, but i cant deal with this 2*k in binomial coefficient. About context: i have solved exercise in which i should find expectation of games number. Sums of this kind is in my answer. Also i see that wolfram calculating this sums, but i never calculated sums of this kind. Or i simply dont see something. – Mark Tuliy May 09 '18 at 07:59
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    "I've tried to use generating functions" OK, this is excellent, but please show in details what you did (which failed) to use GF. – Did May 09 '18 at 08:08
  • @Did Actually i dont know any formulas for sums with those type of binomial coefficient. That was the fail. If u could help with this formula than i think my problem with this approach will be solved. – Mark Tuliy May 09 '18 at 08:12
  • Since you said the factor $2k$ was a problem... Hint: What would be the sum $$\sum_{k=1}^{\infty} C^{k+1}_{2k}x^{2k}$$ say, for $|x|$ small enough? – Did May 09 '18 at 08:17
  • @MarkTuliy If ever you need some inspiration about how to use GF to evaluate similar sums of series... you might want to consult this old answer of mine (but go directly to the evaluation of $A'(x)$ there). – Did May 09 '18 at 08:25
  • @Did, Thanks for this answer – Mark Tuliy May 09 '18 at 08:29
  • You are welcome. But beware that this is only a start, now one needs to extend the method to your case... so if you meet difficulties doing that, please say so. Another point: your post would have been a better contribution to the site if you had mentioned its probabilistic motivation, namely, the probabilistic quantity the sum of the series represent. 'Cause at present, one could think the sum popped out of thin air and that estimating it has no relevance whatsoever... – Did May 09 '18 at 08:31
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    @Did My task was: compute the expectation of the rounds number, if we have k coins in the start, during each round with probability p we lose 1 coin, and with probability 1-p we gain 1 coin. Game is over when we have 0 coins or 6 coins. k = 2 for example. First i've tried recurrence, but i dont know any norming conditions for this.So i get answer as 2 sums of this kind. Probably this exercises has simpler solution, so if u have any ideas i will appreciate it. – Mark Tuliy May 09 '18 at 08:50
  • CAS says: $\sum _{k=1}^{\infty } 2 k \binom{2 k}{k+1} p^{k-1} (1-p)^{k+1}=\frac{-1+\sqrt{(-1+2 p)^2}+2 (-1+p) p \left(-3-2 (-1+p) p+2 \sqrt{(-1+2 p)^2}\right)}{\left((1-2 p)^2\right)^{3/2} p^2}$ for $0<p<1$ – Mariusz Iwaniuk May 09 '18 at 09:06
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    "My task was..." This is much simpler: the expected number of rounds r(k) starting from k solves r(k) = 1 + (1-p) r(k+1) + p r(k-1) for every k from 1 to n - 1, with r(0) = r(n) = 0. Hence s(k) = r(k) - c k for some well chosen c solves s(k) = (1-p) s(k+1) + p s(k-1) and you are done. – Did May 09 '18 at 09:08
  • @Did, oh i miss this 1. Without it i couldnt solve recursion. Thanks) – Mark Tuliy May 09 '18 at 09:24
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    OK. If you reach a solution, do not hesitate to write it down in details and to post the result below as an answer (and even, after a while, to accept it, so that the question has an answer). – Did May 09 '18 at 09:33

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