Sum of the series: $\sum_{n=0}^{\infty} \left(\frac{1}{ 4}\right)^{\!n/2}$

Question

Consider the infinite series $\; \displaystyle \sum_{n=0}^{\infty} \left(\frac{1}{ 4}\right)^{\!n/2}.$

Find the sum of this series.

Formula for calculating sum of the series is:

$$\frac{a}{1 - r}$$

Where Riemann sum must either look as:

$\sum_{n=0}^{\infty} a \cdot r^n\;\;$ or $\;\;\; \sum_{n=1}^{\infty} a \cdot r^{n-1}$

So one way to figure out this problem is to rewrite the given Riemann sum as two sums, one for even and the other for odd n.

Edit: Kind of like this:

$\sum (\frac{1}{4})^n + \sum \frac{1}{2} \cdot (\frac{1}{4})^{\frac{2n}{2}} $

For even, standard is 2k, for odd just simply 2k + 1.

In my case there's fraction involved in the exponent, so I get:

$\sum (\frac{1}{4})^{2n}{2} + \sum (\frac{1}{4})^{ \frac{2n +1}{2}}$ ->

$\sum (\frac{1}{4})^{2n}{2} + \sum (\frac{1}{4})^{ \frac{2n}{2} + \frac{1}{2}}$ ->

$\sum (\frac{1}{4})^n + \sum (\frac{1}{4})^{ \frac{2n}{2}} \cdot (\frac{1}{4})^{ \frac{1}{2}} $

Answer 1

Alternately use index laws:

$$\sum_{n=0}^{\infty} \left(\frac{1}{ 4}\right)^{\frac{n}{2}}=\sum_{n=0}^{\infty} \left(\left(\frac{1}{ 4}\right)^{\frac{1}{2}}\right)^n$$ $$=\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n$$

Answer 2

I think it is $$\sum (\frac{1}{4})^{\frac{n}{2}}=\sum (\frac{1}{4})^{\frac{2n}{2}}+\sum (\frac{1}{4})^{\frac{2n+1}{2}}=\sum (\frac{1}{4})^{\frac{2n}{2}}+\sum (\frac{1}{4})^{\frac{2n}{2}}\times \frac{1}{2}$$

Answer 3

Hint: $$\sum_{n=0}^{\infty} \left(\frac{1}{ 4}\right)^{\!n/2}=(1/4)^{n/2} = 1/2^n$$ Now just use the sum to infinite geometric series.