MHB Compute Two Series: Summation Notation

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The discussion focuses on computing two series using summation notation. The first series converges to \(1 - \sqrt{\frac{1}{2}}\), derived from the expansion of \((1+x)^{-\frac{1}{2}}\). The second series is more complex, involving factorials and the gamma function, ultimately leading to the result \(1 + \frac{\pi}{2}\). The participants utilize properties of arcsine and various mathematical transformations to derive these results. The conversation highlights the intricacies of series summation and the application of advanced mathematical concepts.
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Compute the following series:

$$\frac{1}{2} - \frac{{1 \times 3}}{{2 \times 4}} + \frac{{1 \times 3 \times 5}}{{2 \times 4 \times 6}} \mp \cdots ,$$ $$1 + \frac{{1 \times 2}}{{1 \times 3}} + \frac{{1 \times 2 \times 3}}{{1 \times 3 \times 5}} + \frac{{1 \times 2 \times 3 \times 4}}{{1 \times 3 \times 5 \times 7}} + \cdots .$$
 
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Krizalid said:
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:
Yes, that's correct.
Second series is harder though.
 
I think that the second problem has some thing to do with \( \displaystyle \arcsin(x)=\sum_{0}^{\infty}\frac{(2n)!}{4^n (2n+1) (n!)^2}x^{2n+1}\).
 
Krizalid said:
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:
Krizalid said:
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:

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