Infinite Trigonometric Sums Part 1

The required sum is

$S=\sum \limits_{n\mathop=1}^{\infty}\frac{(-1)^{n-1}n}{2^n} =\frac{1}{2}\sum \limits_{n\mathop=1}^{\infty}n\left(-\frac{1}{2}\right)^{n-1}$

By differentiating the formula $\sum \limits_{n\mathop=1}^{\infty}x^n=\frac{x}{1-x}$ we find

$\sum \limits_{n\mathop=1}^{\infty}nx^{n-1}=\frac{1}{(x-1)^2}$

Substituting $x=-\frac{1}{2}$ we can deduce

$S=\frac{2}{9}$ so $a^2+b^2=\boxed{85}$ .

A slightly different way than Oliver:

The required sum is $S=\displaystyle\sum_{i=1}^{\infty}-i\left(-\dfrac{1}{2}\right)^i$

Note that $-\dfrac{1}{2}S=\displaystyle\sum_{i=1}^{\infty}-i\left(-\dfrac{1}{2}\right)^{i+1}$

Subtracting this second equation from the first, we arrive at the conclusion that $\begin{aligned}\dfrac{3}{2}S&=\left(\displaystyle\sum_{i=1}^{\infty}-i\left(-\dfrac{1}{2}\right)^i\right)-\left(\displaystyle\sum_{i=1}^{\infty}-i\left(-\dfrac{1}{2}\right)^{i+1}\right)\\ &= \displaystyle\sum_{i=1}^{\infty}-\left(-\dfrac{1}{2}\right)^i\\ &= \dfrac{\frac{1}{2}}{1-(-\frac{1}{2})}\\ &= \dfrac{1}{3}\end{aligned}$

Since $\dfrac{3}{2}S=\dfrac{1}{3}$ , then $S=\dfrac{2}{9}$ and our desired answer is $2+9=\boxed{11}$ .

Clearly $\sin{\pi / 6} = \frac{1}{2}$ , so substitute that into the formula.

We have an infinite geometric series of infinite geometric series. In particular,

$S_1 = \frac 12 - \frac 14 + \frac 18 - \dots = \frac{1/2}{3/2}$

$S_2 = - \frac 14 + \frac 18 - \dots = -\frac{1/4}{3/2}$

and so forth.

Thus the required sum is $S = \frac23 \left(\frac 12 - \frac 14 + \frac 18 \dots \right) = \frac23 \left( \frac{1/2}{3/2} \right) = \frac 29$