Summation of an infinite series $\frac{x}{4^x}$

Question

I've found by the ratio test that the following series is convergent, however I am struggling to find the value of the sum. Could anyone suggest where to go from here to find the sum?

$\sum_{i=0}^\infty \frac{i}{4^i} = $

Thanks in advance.

PS - This is the answer via Wolfram Alpha

Answer 1

METHODOLOGY $1$: With Calculus

Let $f(x)=\sum_{n=0}^\infty x^n$. For $|x|<1$, the geometric series converges to $f(x)=\frac{1}{1-x}$.

Note that the series $\sum_{n=0}^\infty nx^{n-1}$, formed by differentiating term-by-term, converges uniformly for all $|x|\le r<1$.

Since the series representation for $f(x)$ converges, then we have $f'(x)=\frac{1}{(1-x)^2}=\sum_{n=0}^\infty nx^{n-1}$ for all $|x|<1$. Hence, we have $\sum_{n=0}^\infty nx^n=\frac{x}{(1-x)^2}$

Now, letting $x=\frac14$, we find

$$\sum_{n=0}^\infty \frac{n}{4^n}=\frac{1/4}{(1-(1/4))^2}=\frac{4}{9}$$

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METHODOLOGY $2$: Without Calculus

Note that we can write

$$\begin{align} \sum_{n=0}^\infty nx^n&=\sum_{n=1}^\infty x^n\sum_{m=1}^{n}(1)\\\\ &=\sum_{m=1}^\infty \sum_{n=m}^{\infty}x^n\\\\ &=\sum_{m=1}^\infty \frac{x^m}{1-x} \\\\ &=\frac{x}{(1-x)^2} \end{align}$$

Answer 2

Hint:
Let's integrate $$\int \frac{x}{4^x}dx=\frac{x}{4^x\log 4}-\frac{1}{(\log4)^24^x}$$ $$ \implies\frac{d}{dx}\left(\frac{x}{4^x\log 4}-\frac{1}{(\log 4)^24^x}\right)=\frac{x}{4^x}$$ if $f(x)=\frac{x}{4^x} $ and $g(x)=\sum_{x=0}^\infty \frac{x}{4^x}$ then: $$g(x)=\frac{g'(x)}{\log 4}-\frac{1}{(\log4)^2}\sum_{x=0}^\infty\frac{1}{4^x}$$ Now it is a differential equation. Solve it.