Find the sum of the series $\sum_{k=0}^\infty \frac{1}{2^{k+3}}$

Question

Find the sum of the series $$\sum_{k=0}^\infty \frac{1}{2^{k+3}}$$

I'm certain the way to go about it is using a the rules regarding a geometric series.

Since we know that $$\sum_{k=0}^\infty \frac{1}{2^{k+3}} = $$ $$\sum_{k=0}^\infty 2^{-k-3} $$ I believe all I know to do is isolate the exponent to k so I can run the geometric series and determine if it is convergent or not.

So I think my problem at this stage is really just algebra. How would isoate the exponent to just "k" rather than (-k-3)

Answer 1

Hint: ${}$${}$ $$2^{k+3}=2^k 2^3$$

Answer 2

$\sum_{k=0}^\infty \frac{1}{2^{k+3}}=\frac{1}{8}\sum_{k=0}^{\infty}2^{-k}=\frac{1}{8}\frac{1}{1-\frac{1}{2}}=\frac{1}{4}$ by geometric series.

Answer 3

It's $$\frac{\frac{1}{8}}{1-\frac{1}{2}}=\frac{1}{4}.$$

Answer 4

Noting that $\frac{1}{2^m} = \left(\frac{1}{2}\right)^m$, when $k = 0$, the first term is just $\frac{1}{8}$, so the sum is equivalent to

$$\sum_{k=0}^\infty \frac{1}{8}\cdot\left(\frac{1}{2}\right)^k = \sum_{k=1}^\infty \frac{1}{8}\cdot\left(\frac{1}{2}\right)^{k-1}$$

Since the common ratio satisfies $\vert r\vert < 1$, as $r = \frac{1}{2}$, the infinite sum converges to $\frac{\frac{1}{8}}{1-\frac{1}{2}} = \frac{1}{4}$.