3

During one exposition of a demonstration, the professor shows the following identity:

$$\sum_{m = 0}^{\infty} \frac{(-1)^m \binom{n}{m}}{(2m+1)} = \frac{(2n)!!}{(2n+1)!!}$$

I've been attempting some identities however with no success (such as Pascal and the definition of binomial).

obs.: $4!! = 4 \cdot2$, $6!! = 6 \cdot4 \cdot 2$, $5!! = 5 \cdot3 \cdot 1$.

Any proof would be great to clear everything.

SimonCello94
  • 343
João Víctor Melo
  • 436
  • As @CHAMSI has shown,$\sum_{m=0}^{n}{\binom{n}{m}\frac{(-1)^{m}}{2m+1}}=\int_{0}^{1}{(1-x^{2})^{n},dx}$, then$$=\int_0^1(1-x)^n(1+x)^n=2^{2n}2\int_0^{1/2}t^n(1-t)^ndt=2^{2n}\int_0^{1}t^n(1-t)^ndt=2^{2n}B(n+1;n+1)$$ On the other hand, $$\frac{(2n)!!}{(2n+1)!!}=\frac{(2n)!!(2n)!!}{(2n+1)!!(2n)!!}=2^{2n}\frac{(n!)^2}{(2n+1)!}=2^{2n}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}=2^{2n}B(n+1;n+1)$$ – Svyatoslav Dec 29 '21 at 16:41
  • What is this function $B(n+1;n+1)$? – João Víctor Melo Dec 29 '21 at 16:52
  • 1
    Related. The accepted answer proves this is true. https://math.stackexchange.com/questions/1528824/is-the-numerator-of-sum-k-0n-frac-1k2k1-binomnk-a-power-of-2 – Thomas Andrews Dec 29 '21 at 16:56
  • Why could CHAMSI in the answer below exchange the integral with the sum to get $(1-x^2)^n$? – João Víctor Melo Dec 29 '21 at 17:02
  • Beta-function $B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ https://en.wikipedia.org/wiki/Beta_function – Svyatoslav Dec 29 '21 at 17:03
  • @dynamicalsystems Since the integral is linear and the sum is finite, we can exchange the sum and the integral, then apply the binomial theorem. – CHAMSI Dec 29 '21 at 17:35
  • What is the trick to divide the integral in two? – João Víctor Melo Dec 29 '21 at 18:28
  • In this answer. in the derivation, there appears $$\sum_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}{k}=\frac{2^nn!}{1\cdot3\cdots(2n+1)}$$ and the latter fraction is $\frac{(2n)!!}{(2n+1)!!}$ – robjohn Dec 30 '21 at 02:28
  • Do any of the answers to this question answer your question? – robjohn Dec 30 '21 at 02:34

1 Answers1

6

Your infinite sum is actually finite, because $ \binom{n}{m}=0, \left(\forall m>n\right) $.

Let $ n\in\mathbb{N} $, denoting $ u_{n}=\sum\limits_{m=0}^{n}{\binom{n}{m}\frac{\left(-1\right)^{m}}{2m+1}} $, we have : \begin{aligned}u_{n}=\sum_{m=0}^{n}{\binom{n}{m}\frac{\left(-1\right)^{m}}{2m+1}}&=\sum_{m=0}^{n}{\binom{n}{m}\left(-1\right)^{m}\int_{0}^{1}{x^{2m}\,\mathrm{d}x}}\\ &=\int_{0}^{1}{\left(1-x^{2}\right)^{n}\,\mathrm{d}x}\\ &=\left[x\left(1-x^{2}\right)^{n}\right]_{0}^{1}+2n\int_{0}^{1}{x^{2}\left(1-x^{2}\right)^{n-1}\,\mathrm{d}x}\\ &=2n\int_{0}^{1}{\left(1-\left(1-x^{2}\right)\right)\left(1-x^{2}\right)^{n-1}\,\mathrm{d}x}\\ u_{n}&=2n\left(u_{n-1}-u_{n}\right)\\ \iff u_{n}&=\frac{2n}{2n+1}u_{n-1}\\&=\prod_{k=1}^{n}{\frac{2k}{2k+1}}u_{0}=\frac{\left(2n\right)!!}{\left(2n+1\right)!!}\end{aligned}

CHAMSI
  • 9,222