How do I find the sum of the series: (sum. n=1, infinity) [2^(n+1)]/pi^n
This looks to be a geometric series, and you are trying to find the sum as it goes to infinite. So lets break it down
$\displaystyle \sum_{n=1}^{\infty}\frac{2^{n+1}}{\pi^n}$
Now If you can remember properties of exponents we can transform the numerator to help us
$\displaystyle \sum_{n=1}^{\infty}\frac{2^{n}\cdot2^1}{\pi^n}$
We can pull that out as a constant multiplicative of this geometric series
$\displaystyle \sum_{n=1}^{\infty}2\cdot\frac{2^{n}}{\pi^n}$
$\displaystyle \sum_{n=1}^{\infty}2\cdot\frac{2^{n}}{\pi^n}$
We can use another property of exponents here
$\displaystyle \sum_{n=1}^{\infty}2\cdot(\frac{2}{\pi})^n$
Now the rest is easy, Our constant multiple lets call it a which is the first term in the series, is $\displaystyle a = 2\cdot\frac{2}{\pi}$
and your common ration r is $\displaystyle r=\frac{2}{\pi}$, now if we recall the formula for a sum of the geometric series as it goes to infinity
$\displaystyle \frac{a}{1-r}$
You should be able to get the rest
Hello, schnek!
We have: .$\displaystyle S \;=\;\frac{2^2}{\pi} + \frac{2^3}{\pi^2} + \frac{2^4}{\pi^3} + \hdots$Find the sum of the series: .$\displaystyle S \;=\;\sum^{\infty}_{n=1} \frac{2^{n+1}}{\pi^n} $
This is a geometric series with first term $\displaystyle a = \frac{4}{\pi}$ and common ratio $\displaystyle r = \frac{2}{\pi}$
The sum is: .$\displaystyle S \;=\;\frac{a}{1-r} \;=\;\frac{\frac{4}{\pi}}{1 - \frac{2}{\pi}} \;=\;\frac{4}{\pi-2}$