# Suggest a tricky method for this problem

Find the value of:

$$1+ \biggl(\frac{1}{10}\biggr)^2 + \frac{1 \cdot 3}{1 \cdot 2} \biggl(\frac{1}{10}\biggr)^4 + \frac{1\cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \biggl(\frac{1}{10}\biggr)^6 + \cdots$$

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Hint: What is the generalized binomial expansion of $\left( 1-2 \times \left(\frac{1}{10}\right)^2 \right) ^{-\frac{1}{2}}$?

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And how you derive this?Precisely I am interested in your method,I have a solution but it's not feasible under the 1 mint rule. – Quixotic Nov 29 '10 at 14:12
@Debanjan: Well I just looked at the terms of the series and the geometrically increasing $(\frac{1}{10})^2$ and the product of the first $r$ natural numbers in the denominator suggested binomial theorem. The fact that numerator had only odd numbers suggested that the exponent must $\pm \frac{1}{2}$. Then I just looked at the signs in both expansions. This still resulted in an excess of $2^r$ in each term, so there must be a $2$ in the original binomial. To prove these are the same series, I just need to show that they have the same expansion which is done using generalized binomial expansion – Timothy Wagner Nov 29 '10 at 14:23
(contd): A lot of these series based questions can be solved by comparing the given series to a well known one. Of course, it comes with the proviso that you know some standard series expansions. This article on wikipedia should help with some well known series expansions en.wikipedia.org/wiki/Taylor_series – Timothy Wagner Nov 29 '10 at 15:00
Indeed it was a beautiful solution, I don't know why I forgot about this one, anyways +1 and accepted :) Thanks again. – Quixotic Nov 25 '11 at 15:40

Note that we can rewrite the series as $$\displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot \bigg(\frac{1}{10}\bigg)^{2n}$$ which is exactly $$\displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot x^{2n}$$ evaluated at $x = \frac{1}{10}$. It is easy to see that this sum has radius of convergence $\frac{1}{\sqrt{2}}$, and so converges absolutely for $x = \frac{1}{10}$.

Also note that this sum is exactly the Taylor series for $\frac{1}{\sqrt{1-2x^2}}$ centered at $x = 0$. Thus $$\displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot \bigg(\frac{1}{10}\bigg)^{2n} = \frac{10}{7\sqrt{2}}$$

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This can also be done using Wallis' product.

For instance see a similar problem: Summing the series $\frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$

and

Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$

Use

$$\displaystyle \dfrac{2}{\pi} \int_{0}^{\pi/2} \sin^{2n} x \ \text{dx} = \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} = \dfrac{1}{2^n} \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{1 \cdot 2 \cdot 3 \cdots n}$$

Thus

$$\displaystyle \dfrac{2}{\pi} \int_{0}^{\pi/2} \left(\dfrac{2\sin^2 x}{100}\right)^n \ \text{dx} = \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{1 \cdot 2 \cdot 3 \cdots n} \dfrac{1}{10^{2n}}$$

Thus your sum is

$$\displaystyle \sum_{n=0}^{\infty} \dfrac{2}{\pi} \int_{0}^{\pi/2} \left(\dfrac{2\sin^2 x}{100}\right)^n \ \text{dx} = \dfrac{2}{\pi} \int_{0}^{\pi/2} \sum_{n=0}^{\infty} \left(\dfrac{2\sin^2 x}{100}\right)^n \ \text{dx}$$

$$\displaystyle = \dfrac{2}{\pi}\int_{0}^{\pi/2} \dfrac{100}{100 - 2\sin^2x} \ \text{dx} = \dfrac{2}{\pi} \dfrac{5\pi}{7\sqrt{2}} = \dfrac{10}{7\sqrt{2}}$$

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Wallis, Wallis, always Wallis... :D – J. M. Nov 29 '10 at 13:06
@J.M: Yeah! :-D – Aryabhata Nov 29 '10 at 16:14