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Compute two series

Krizalid

Active member
Feb 9, 2012
118
Compute the following series:

$$\frac{1}{2} - \frac{{1 \times 3}}{{2 \times 4}} + \frac{{1 \times 3 \times 5}}{{2 \times 4 \times 6}} \mp \cdots ,$$ $$1 + \frac{{1 \times 2}}{{1 \times 3}} + \frac{{1 \times 2 \times 3}}{{1 \times 3 \times 5}} + \frac{{1 \times 2 \times 3 \times 4}}{{1 \times 3 \times 5 \times 7}} + \cdots .$$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Compute the following series:

$$\frac{1}{2} - \frac{{1 \times 3}}{{2 \times 4}} + \frac{{1 \times 3 \times 5}}{{2 \times 4 \times 6}} \mp \cdots ,$$
Remembering that is...

$\displaystyle (1+x)^{-\frac{1}{2}}= 1 - \frac{1}{2}\ x + \frac{1\ 3}{2\ 4}\ x^{2} - \frac{1\ 3\ 5}{2\ 4\ 6}\ x^{3} + ...$ (1)

... is...

$\displaystyle \frac{1}{2} - \frac{1\ 3}{2\ 4} + \frac{1\ 3\ 5}{2\ 4\ 6}- ...= 1- \sqrt{\frac{1}{2}}$ (2)

Kind regards

$\chi$ $\sigma$
 
Last edited:

Krizalid

Active member
Feb 9, 2012
118
Yes, that's correct.
Second series is harder though.
 

sbhatnagar

Active member
Jan 27, 2012
95
I think that the second problem has some thing to do with \( \displaystyle \arcsin(x)=\sum_{0}^{\infty}\frac{(2n)!}{4^n (2n+1) (n!)^2}x^{2n+1}\).
 

chisigma

Well-known member
Feb 13, 2012
1,704
Compute the following series:

$$1 + \frac{{1 \times 2}}{{1 \times 3}} + \frac{{1 \times 2 \times 3}}{{1 \times 3 \times 5}} + \frac{{1 \times 2 \times 3 \times 4}}{{1 \times 3 \times 5 \times 7}} + \cdots .$$
The solution of the second series is effectively a little more difficult task!... let's start defining...

$\displaystyle \varphi(x)= \sum_{n=1}^{\infty} \frac{n!}{(2n-1)!!}\ x^{n}$ (1)

... so that is $\displaystyle \sum_{n=1}^{\infty} \frac{n!}{(2n-1)!!}= \varphi(1)$. Introducing the gamma function, taking into account that is...

$\displaystyle (2n-1)!!= \frac{2^{n}}{\sqrt{\pi}}\ \Gamma(n+\frac{1}{2})$ (2)

... the (1) becomes...

$\displaystyle \varphi(x)= \sqrt{\pi} \sum_{n=1}^{\infty} \frac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})}\ (\frac{x}{2})^{n}$ (3)

Now we consult the excellent library of the University of Bonn and here we find...

$\displaystyle \sum_{n=0}^{\infty} \frac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})}\ z^{n}= \frac{1}{\sqrt{\pi}\ (1-z)}\ (1+ \frac{\sqrt{z}\ \sin^{-1} \sqrt{z}}{\sqrt{1-z}}) $ (4)

... so that is...

$\displaystyle \sum_{n=1}^{\infty} \frac{n!}{(2n-1)!!}= 2\ (1+ \sin^{-1} \frac{1}{\sqrt{2}})-1= 1 + \frac{\pi}{2}$

Kind regards

$\chi$ $\sigma$
 
Last edited:

Sherlock

Member
Jan 28, 2012
59
Compute the following series:

$$1 + \frac{{1 \times 2}}{{1 \times 3}} + \frac{{1 \times 2 \times 3}}{{1 \times 3 \times 5}} + \frac{{1 \times 2 \times 3 \times 4}}{{1 \times 3 \times 5 \times 7}} + \cdots .$$
The series can be written as

$\displaystyle \sum_{k \ge 1}\frac{k!}{(2k-1)!!} = \sum_{k \ge 1}\frac{k!^2 2^k}{(2k)!} = \sum_{ k \ge 1}\frac{2^k}{\binom{2k}{k}} $

Consider the series

$\displaystyle \left(\sin^{-1}{x}\right)^2 = \sum_{k \ge 1}\frac{2^{2k-1}x^{2k}}{k^2 \binom{2k}{k}}$

Differentiating both sides gives


$\displaystyle\frac
{2\sin^{-1}{x}}{\sqrt{1-x^2}} = \sum_{k \ge 1}\frac{2^{2k}x^{2k-1}}{k \binom{2k}{k}}$

Rearrange it as

$\displaystyle\frac
{x\sin^{-1}{x}}{\sqrt{1-x^2}} = \sum_{k \ge 1}\frac{2^{2k-1}x^{2k}}{k \binom{2k}{k}}$

Differentiating both sides we get

$\displaystyle \frac
{\sin^{-1}{x}+x\sqrt{1-x^2}}{(1-x^2)\sqrt{1-x^2}} = \sum_{k \ge 1}\frac{2^{2k}x^{2k-1}}{\binom{2k}{k}}$

Rearrange it as

$\displaystyle \frac
{x\sin^{-1}{x}+x^2\sqrt{1-x^2}}{(1-x^2) \sqrt{1-x^2}} = \sum_{k \ge 1}\frac{(2x)^{2k}}{\binom{2k}{k}}$

Put $x = \frac{1}{\sqrt{2}}$, then:

$\displaystyle 1+\frac{\pi}{2} = \sum_{k \ge 1}\frac{2^k}{\binom{2k}{k}}.$