Infinite sum of sequence

Question

I was just curious to know how to find the infinite sum of the below sequence.

$$\frac{1}{4}+\frac{2}{4^2}+\frac{1}{4^3}+\frac{2}{4^4}+\frac{1}{4^5}+\frac{2}{4^6}+\frac{1}{4^7}+\dots$$

I know it can be written as two sigmas but I don't know how to find the infinite sum of those.

So please find the sum and explain how you did it step by step

Answer 1

$$\sum_{r=0}\dfrac1{4^{2r+1}}+2\sum_{r=0}\dfrac1{4^{2r}}$$

Now use $$\sum_{n=0}^\infty ar^n=\dfrac a{1-r},\text{for }|r|<1$$

independently for both the summations

Answer 2

Let $\mathcal{S}$ denote the given sum. Then:

\begin{align*} \mathcal{S} &=\sum_{k=0}^{\infty} \frac{1}{4^{2k+1}} + \sum_{k=0}^{\infty} \frac{2}{4^{2k}} \\ &= \frac{1}{4}\sum_{k=0}^{\infty} \frac{1}{4^{2k}} + 2 \sum_{k=0}^{\infty} \frac{1}{4^{2k}}\\ &=\left ( 2+ \frac{1}{4} \right ) \sum_{k=0}^{\infty} \frac{1}{4^{2k}} \\ &= \frac{9}{4}\sum_{k=0}^{\infty} \frac{1}{16^k} \\ &= \frac{9}{4} \frac{1}{1- \frac{1}{16}} \\ &=\frac{9}{4}\cdot \frac{16}{15} \\ &=\frac{36}{15} \end{align*}

since $\sum \limits_{k=0}^{\infty} a^k = \frac{1}{1-a} , \; |a|<1$.

Answer 3

This is more or less the same answer as the others, but it really simplifies finding the sum if you recognize

$$ \dfrac{1}{4} + \dfrac{2}{4^2} + \dfrac{1}{4^3} + \dfrac{2}{4^4} + \dfrac{1}{4^5} + \dfrac{2}{4^6} + \dfrac{1}{4^7} + ...$$

as

$$ \dfrac{1}{4} + \left(\dfrac{1}{4^2} + \dfrac{1}{4^2}\right) + \dfrac{1}{4^3} + \left(\dfrac{1}{4^4} + \dfrac{1}{4^4}\right) + \dfrac{1}{4^5} + \left(\dfrac{1}{4^6} + \dfrac{1}{4^6}\right) + \dfrac{1}{4^7} + ... $$

Rearranging, you get

\begin{align*} &= \left(\dfrac{1}{4} + \dfrac{1}{4^2} + \dfrac{1}{4^3} + \dfrac{1}{4^4} + \dfrac{1}{4^5} + \dfrac{1}{4^6} + \dfrac{1}{4^7} + ... \right) + \left(\dfrac{1}{4^2} + \dfrac{1}{4^4} + \dfrac{1}{4^6} + ...\right) \\ &= \left(\dfrac{1}{4} + \dfrac{1}{4^2} + \dfrac{1}{4^3} + ... \right) + \left(\dfrac{1}{16} + \dfrac{1}{16^2} + \dfrac{1}{16^3} + ...\right) \\ &= \sum_{n=1}^{\infty} {\dfrac{1}{4^n}} + \sum_{n=1}^{\infty} {\dfrac{1}{16^n}} \end{align*}

The simpler elements of the summation help avoid off-by-one errors in the summation index that lead to the wrong answer.