Suggest a tricky method for this problem Find the value of:
$$ 1+ \biggl(\frac{1}{10}\biggr)^2 + \frac{1 \cdot 3}{1 \cdot 2} \biggl(\frac{1}{10}\biggr)^4 + \frac{1\cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \biggl(\frac{1}{10}\biggr)^6 + \cdots  $$
 A: 
Hint: What is the generalized binomial expansion of $\left( 1-2 \times
 \left(\frac{1}{10}\right)^2 \right) ^{-\frac{1}{2}}$?

A: 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
$$ \displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot \bigg(\frac{1}{10}\bigg)^{2n} = \frac{10}{7\sqrt{2}}$$
A: 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
$$\displaystyle  \sum_{n=0}^{\infty} \dfrac{2}{\pi} \int_{0}^{\pi/2} \left(\dfrac{2\sin^2 x}{100}\right)^n \ \text{dx} =  \dfrac{2}{\pi} \int_{0}^{\pi/2} \sum_{n=0}^{\infty} \left(\dfrac{2\sin^2 x}{100}\right)^n  \ \text{dx}$$
$$\displaystyle = \dfrac{2}{\pi}\int_{0}^{\pi/2} \dfrac{100}{100 - 2\sin^2x} \ \text{dx} = \dfrac{2}{\pi} \dfrac{5\pi}{7\sqrt{2}} = \dfrac{10}{7\sqrt{2}}$$
