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I have found the following problem here: https://brilliant.org/problems/intriguing-sum/?group=Km7yEIDGtHDa&ref_id=709399

In the solution a solver directly started with the equation given in the title, (which is slightly different than the sum given in the problem! The computation of the value $ S $ is sought:) $ S = \sum_{n=1}^{\infty} {\binom{2n}{n} \over n* 4^n } $

In particular I wonder, how one can assume per se, that such $ m $ and $ n $ in the first term exist, so that the equation holds $ (1+mx)^n = 1 + \sum_{n=1}^{\infty} {\binom{2n}{n} \over 4^n } x^n $ ; I don't find this clear or trivial.

He continued comparing the coefficients of $x$ and got $ mn = 1/2, m^2 * {n(n+1) \over 2 } = 1.3/2.4 $,

$1/2 $ being the first value of the sum, however I also don't see where $1.3/2.4 $ came from :/

Newton's generalized binomial theorem states: $ {1 \over (1-x)^s } = \sum_{k = 0}^{\infty} {\binom{s-k+1}{k}} $ for a some $ s $, but this doesn't help me yet to get to terms above.

I wished I would understand the solution, but somehow I feel a bit stupid, for not even understanding a solution, if one is presented..

I would be really thankful, if someone could help me out. If further information are required, I add them.

Imago
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1 Answers1

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Let $1+\sum_{r=1}^\infty\binom{2r}rx^r=(1+mx)^n$

$$1+2x+6x^2+\cdots=1+mn(x)+\binom n2m^2 x^2+\cdots$$

Equating the coefficients of $x,mn=2\iff m=2/n$

and equating the coefficients of $x^2,$ $m^2\dfrac{n(n-1)}2=6\iff\dfrac4{n^2}\dfrac{n(n-1)}2=6\iff\dfrac{n-1}n=3\iff n=-1/2$

and subsequently $m=-4$

$$\implies1+\sum_{r=1}^\infty\binom{2r}rx^r=(1-4x)^{-1/2}$$ whose $u(\ge1)$th term is $$\dfrac{(-4x)^u\left(-\dfrac12\right)\left(-\dfrac12-1\right)\cdots\left[-\dfrac12-(u-1)\right]}{u!}$$

$$=\dfrac{x^u[1\cdot3\cdot5\cdots(2u-1)]}{u!}=x^u\binom{2u}u$$

lab bhattacharjee
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