Calculus for a Sum

Here is a solution without calculus.

The sum can be split up into multiple different sums: $\begin{aligned}\sum\limits_{n=1}^\infty\frac{n}{4^n}&=\frac{1}{4}&&+\frac{2}{4^2}&&+\frac{3}{4^3}&&+\frac{4}{4^4}&&+\dots\\\\ &=\frac{1}{4}&&+\frac{1}{4^2}&&+\frac{1}{4^3}&&+\frac{1}{4^4}&&+\dots\\ &&&+\frac{1}{4^2}&&+\frac{1}{4^3}&&+\frac{1}{4^4}&&+\dots\\ &&&&&+\frac{1}{4^3}&&+\frac{1}{4^4}&&+\dots\\ &&&&&&&+\frac{1}{4^4}&&+\dots\\ &&&&&&&&\vdots\end{aligned}$ We can find the value of each of these sums using the formula for the sum of an infinite geometric series: $S=\frac{a}{1-r}\\ \left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\dots\right)+\left(\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+\dots\right)+\left(\frac{1}{4^3}+\frac{1}{4^4}+\frac{1}{4^5}+\dots\right)+\dots\\ =\left(\frac{\frac{1}{4}}{1-\frac{1}{4}}\right)+\left(\frac{\frac{1}{4^2}}{1-\frac{1}{4}}\right)+\left(\frac{\frac{1}{4^3}}{1-\frac{1}{4}}\right)+\dots\\ =\frac{1}{3}+\frac{1}{3\times4}+\frac{1}{3\times4^2}+\dots$ Now we can use the formula again on the new series: $S=\frac{a}{1-r}\\ \begin{aligned}\frac{1}{3}+\frac{1}{3\times4}+\frac{1}{3\times4^2}+\dots&=\frac{\frac{1}{3}}{1-\frac{1}{4}}\\ &=\frac{4}{9}\end{aligned}$ So $a=4$ and $b=9$ , so the final answer is $4+9=\boxed{13}$

Relevant wiki: Geometric Progression Sum

$\begin{aligned} S & = \sum_{n=1}^\infty \frac n{4^n} & \small \color{#3D99F6} \text{Expand the summation} \\ S & = \frac 14 + \frac 2{4^2} + \frac 3{4^3} + \frac 4{4^4} + \cdots & \small \color{#3D99F6} \text{Multiply both sides by }4 \\ 4S & = 1 + \frac 24 + \frac 3{4^2} + \frac 4{4^3} + \cdots \\ 4S - S & =1 + \frac 14 + \frac 1{4^2} + \frac 1{4^3} + \cdots \\ 3S & =\color{#3D99F6} 1 + \frac 14 + \frac 1{4^2} + \frac 1{4^3} + \cdots & \small \color{#3D99F6} \text{Multiply both sides by }4 \\ 12S & = 4 + \color{#3D99F6} 1 + \frac 14 + \frac 1{4^2} + \frac 1{4^3} + \cdots \\ 12S & = 4 + \color{#3D99F6} 3S \\ \implies S & = \frac 49 \end{aligned}$

Therefore, $a+b=4+9 = \boxed{13}$

If you try to fumble with the equation without using calculus, you'll have to be very creative. First, make note of the "parent" sum $\sum_{n=0}^\infty x^n$ and how it is equal to $\frac{1}{1-x}$ for $|x|<1$ . We can observe that we have something similar to this occurring if we split the fraction from $\frac{n}{4^n}$ to $n \cdot (\frac{1}{4})^{n}$ , where $\frac{1}{4}$ is very similar to $x$ , and has an absolute value less than 1. As mentioned before, this sum is equivalent to the formula for an infinite geometric sum: $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ .

If you take the derivative of both sides: $\frac{d}{dx} \sum_{n=0}^\infty x^n =\frac{d}{dx} \frac{1}{1-x}$

Since the first term of the sum on the left is $x^0$ = 1, taking the derivative will be zero. Other than that, the derivative will be $1+2x+3x^2+\ldots$ which can be rewritten as $\sum_{n=1}^\infty n \cdot x^{n-1}$ .

The right hand side rewrites as $\frac{d}{dx}(1-x)^{-1}$ which is (using the chain rule) $-1(x-1)^{-2}\cdot -1=\frac{1}{(1-x)^2}$ The next step is to plug $\frac{1}{4}$ in for $x$ on the right hand side: $\frac{1}{(1-\frac{1}{4})^2}=\frac{1}{(\frac{3}{4})^2}=\frac{1}{\frac{9}{16}}=\frac{16}{9}$ and the left hand side: $\sum_{n=1}^\infty n \cdot \frac{1}{4^{n-1}}$

The left side has now been shown to equal $\frac{16}{9}$ , now the only thing left is to figure out what to do with the left.

If you recall that $a\cdot \sum b = \sum a\cdot b$ for any unchanging value of a. If you pick $a$ to be $\frac{1}{4}$ , then $\frac{1}{4}\cdot \sum_{n=1}^\infty \frac{n}{4^{n-1}}$ becomes $\sum_{n=1}^\infty \frac{1}{4}\cdot \frac{n}{4^{n-1}}$ or $\sum_{n=1}^\infty \frac{n}{4^n}$ and you also have to multiply the right side by $\frac{1}{4}$ which equals $\frac{4}{9}$ in the simplest form $\frac{a}{b}$ , $4+9=\boxed{13}$