# Thread: [SOLVED] Series - Geometric Sum Formula

1. ## [SOLVED] Series - Geometric Sum Formula

Problem:
Find a formula for
a.
$\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{( 1+x^2)^n}$
b. Also, show that if $a \neq 0$, then
$\frac{1}{a}=1+(1-a)+(1-a)^2+\frac{(1-a)^3}{a}$
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a. By the Geometric Sum Formula,
$\sum_{k=0}^{n}r^k = \frac{1-r^{n+1}}{1-r}$ (1)

Since $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{( 1+x^2)^n}= \sum_{k=0}^{n} \left( \frac{1}{1+x^2} \right)^n = \sum_{k=0}^{n} \frac{1}{(1+x^2)^n}$ (2)

Let $r= \frac{1}{1+x^2}$, then substitute r back into (1) yields

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

Before I simplify this fraction, this formula doesn't seem right to me.
Suppose n=2 and x=2, then the equation (2) becomes

$\frac{1}{1+2^2}+\frac{1}{(1+2^2)^2}= \frac{1}{5} + \frac{1}{25} = \frac{6}{25}$

However, if you plug n =2 and x = 2 into equation (3) , then it becomes
$\frac{1- \left( \frac{1}{5}\right)^3}{1-\frac{1}{5}} = \frac{1 - \frac{1}{125}}{1-\frac{1}{5}} = \frac{31}{25}$

But $\frac{6}{25} \neq \frac{31}{25}$

(b) Since $1+r+r^2+...+r^n= \frac{1-r^{n+1}}{1-r}$

$\implies [ 1 + (1-a) + (1-a)^2 ] + \frac{(1-a)^3}{a}$

$\implies \frac{1-(1-a)^3}{1-(1-a)} + \frac{(1-a)^3}{a}$

$\implies \frac{1-(1-a)^3}{a} + \frac{(1-a)^3}{a}$
$= \frac{1}{a}$

I'm unsure what I am doing incorrectly for part (a).

2. The formula for the first part should be

$\dfrac{\dfrac{1}{1+x^2}\left[1-\left(\dfrac{1}{1+x^2}\right)^n\right]}{1-\dfrac{1}{1+x^2}}$

3. Originally Posted by JaneBennet
The formula for the first part should be

$\dfrac{\dfrac{1}{1+x^2}\left[1-\left(\dfrac{1}{1+x^2}\right)^n\right]}{1-\dfrac{1}{1+x^2}}$
Thank you Jane, this is because $\sum_{k=0}^{n} \frac{1}{(1+x^2)^n}$
k actually starts at 1.

After some algebra, I get $\frac{1}{x^2} \left( 1 - \frac{1}{(1-x^2)^n}\right)$