Summation #7

Relevant wiki: Maclaurin Series

Similar solution with @Guilherme Niedu 's, different presentation.

$\begin{aligned} S & = \sum_{k=0}^\infty \frac {(-1)^k \color{#3D99F6}\left(\frac 14\right)^{k+1}}{{\color{#D61F06}2^{2k}}(2k+1)!} \\ & = \sum_{k=0}^\infty \frac {(-1)^k}{(2k+1)!} \cdot \frac 1{{\color{#D61F06}4^k}\cdot \color{#3D99F6}4^{k+1}} \\ & = \sum_{k=0}^\infty \frac {(-1)^k}{(2k+1)!} \left(\frac 14\right)^{2k+1} & \small \color{#3D99F6} \text{Maclaurin series }\sin x = \sum_{n=0}^\infty \frac {(-1)^n x^{2n+1}}{(2n+1)!} \\ & = \sin \left(\frac 14\right) \end{aligned}$

Therefore, $a+b = 1+3 = \boxed 4$ .

$\large \displaystyle S = \sum_{k=0}^{\infty} \frac{ (-1)^k \left(\frac14 \right) ^{k+1} }{\color{#20A900}2^{2k} \color{#333333} (2k+1)!}$

$\large \displaystyle S = \sum_{k=0}^{\infty} \frac{ (-1)^k \left(\frac14 \right) ^{k+1} }{(2k+1)!} \color{#20A900} \left(\frac14 \right) ^{k}$

$\large \displaystyle S = \sum_{k=0}^{\infty} \frac{ (-1)^k \left(\frac14 \right) ^{2k+1} }{(2k+1)!}$

Which is the Maclaurin series for the sine function.

So:

$\color{#3D99F6} \large \displaystyle S = \sin \left ( \frac14 \right ) , a = 1, b = 4, \boxed{\large \displaystyle a+b=5}$

The given summation after expansion gives us the following infinite series -

$\begin{aligned} S &= \large\sum_{k=0}^\infty\frac{(-1)^{k}\left(\frac{1}{4}\right)^{k+1}}{2^{2k}(2k+1)!} \\ \\ &= \frac{1}{2^2} - \frac{1}{2^6{3}!} + \frac{1}{2^{10}{5}!} - \frac{1}{2^{14}{7}!} \dots (1) \\ \end{aligned}$

And this looks exactly like the Taylor Series Expansion of $sinx$ centered at $x=0$ , which is otherwise referred to as the Maclaurin Series expansion of $sinx$ .

The Maclaurin Series Expansion of $sinx$ is

$\large{sinx = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \dots (2)}$

Comparing equations $(1)$ and $(2)$ we get $\large{x = \frac{1}{2^2} = \frac{1}{4}}$

So $\large{sinx = sin(\frac{1}{4}) = sin(\frac{a}{b})}$

Hence $a+b = 5$

For more info on Taylor Series refer this . Anyone will fall in love with Taylor Series :-)