Convergent or divergent? Find sum if convergent. (problem 2)

**superduper1** · November 18th 2012, 05:41 AM

$\sum\limits_{i=1}^{infty} \frac{(-2)^{n+4}}{\pi^n}$

**Soroban** · November 18th 2012, 06:29 AM

Hello, superduper1!

$\sum\limits_{n=1}^{\infty} \frac{(\text{-}2)^{n+4}}{\pi^n}$

This is a geometric series.
. . The first term is: . $a \;=\;\frac{(\text{-}2)^5}{\pi^1} \;=\;\text{-}\frac{32}{\pi}$
. . The common ratio is: . $r \;=\;\text{-}\frac{2}{\pi}$

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

. . $S \;=\;\dfrac{\text{-}\frac{32}{\pi}}{1-(\text{-}\frac{2}{\pi})} \;=\; \dfrac{\text{-}\frac{32}{\pi}}{\frac{\pi+2}{\pi}} \;=\; \frac{\text{-}32}{\pi+2}$

**superduper1** · November 18th 2012, 06:38 AM

How did you get the common ratio?

**topsquark** · November 18th 2012, 08:37 AM

Originally Posted by superduper1

How did you get the common ratio?

r is the ratio of the n+1th term by the nth term. In this case:
$r = \frac{ \frac{(-2)^{(n + 4) + 1}}{\pi ^ {n + 1}} }{\frac{(-2)^{n + 4}}{\pi ^ n} }$

**skeeter** · November 18th 2012, 09:09 AM

$\sum_{i=1}^{\infty} \frac{(-2)^{n+4}}{\pi^n} = 16 \sum_{i=1}^{\infty} \left(\frac{-2}{\pi}\right)^n = 16 \cdot \frac{\frac{-2}{\pi}}{1 + \frac{2}{\pi}} = \frac{-32}{\pi+2}$

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Convergent or divergent? Find sum if convergent. (problem 2)

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