geometric series

**TAG16** · March 17th 2009, 12:45 PM

Problem 1

n=0 to infinity (-1/2)^n (x-3)^n

How would you find the value of x for which the given geometric series converges and how would you find the sum of the series(as a function of x) for the value of x?

Problem 2

How would you find the values of b for which

1+ e^b+ e^2b+ e^3b+.....=9

**Grandad** · March 17th 2009, 01:09 PM

Hello TAG16

Originally Posted by TAG16

Problem 1

n=0 to infinity (-1/2)^n (x-3)^n

How would you find the value of x for which the given geometric series converges and how would you find the sum of the series(as a function of x) for the value of x?

$(-\tfrac{1}{2})^n(x-3)^n = (-\tfrac{1}{2}(x-3))^n =\left(\frac{3-x}{2}\right)^n$

So the series converges if $-1 < \frac{3-x}{2}<1$

$\Rightarrow -2<3-x<2$

$\Rightarrow -5<-x<-1$

$\Rightarrow 5>x>1$

The sum to infinity $S= \frac{a}{1-r}$ , where $a = 1$ and $r = \frac{3-x}{2}$

$\Rightarrow S =$ ... you can finish it now.

Problem 2

How would you find the values of b for which

1+ e^b+ e^2b+ e^3b+.....=9

This is a GP where $a = 1$ and $r = e^b$ . So the sum to infinity $S= \frac{a}{1-r} = \frac{1}{1 - e^b} = 9$

Solve this equation for $b$ . (The answer will involve $\text{ln}$ .)

Grandad

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