# Thread: Sum of the series

1. ## Sum of the series

How do I find the sum of the series: (sum. n=1, infinity) [2^(n+1)]/pi^n

2. Originally Posted by schnek
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

3. One small caveat...the sum starts at 1, so $\displaystyle a$ should be $\displaystyle 2\cdot\frac{2}{\pi}$.

4. Hello, schnek!

Find the sum of the series: .$\displaystyle S \;=\;\sum^{\infty}_{n=1} \frac{2^{n+1}}{\pi^n}$
We have: .$\displaystyle S \;=\;\frac{2^2}{\pi} + \frac{2^3}{\pi^2} + \frac{2^4}{\pi^3} + \hdots$

This is a geometric series with first term $\displaystyle a = \frac{4}{\pi}$ and common ratio $\displaystyle r = \frac{2}{\pi}$

The sum is: .$\displaystyle S \;=\;\frac{a}{1-r} \;=\;\frac{\frac{4}{\pi}}{1 - \frac{2}{\pi}} \;=\;\frac{4}{\pi-2}$

5. Aw, I knew I was gonna make a careless mistake, that the numerator is the first term in the series. , Thanks Sox, learned from this mistake