# Geometric Progression

1. Apr 8, 2009

### latentcorpse

what is the nth partial sum of

$1-x+x^2-x^3+..$

i don't understand why i can't do this?

i have $\sum_{k=1}^{n} ar^k=\frac{a(r^{n+1}-r)}{r-1}$

ok but then when i sub in i get messed up i get $\frac{(-x)^{n+1}+x}{-(1+x)}$ is there any way of simplifying this? also when i set n=1 i get that the sum is -x when it should be 1-x???

2. Apr 8, 2009

### e(ho0n3

Do you know how to derive the nth partial sum of 1 + x + x^2 + ... ?

3. Apr 8, 2009

### latentcorpse

off the top of my head isn't it something like

$(1-x)(1+x+x^2+...+x^n)=1-x^{n+1} \Rightarrow 1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}$

4. Apr 8, 2009

### latentcorpse

also, another thing i'm having trouble seeing is why $\sum \frac{1}{n+1}$ diverges.
my initial instinct was to use the copmarison test with $\sum \frac{1}{n}$ as this diverges.

so if i can show $\frac{1}{1+n} \geq \frac{1}{n}$ then all is well but in fact $\frac{1}{n} \geq \frac{1}{1+n}$ e.g. set n=1 to see

what's going on here?

5. Apr 8, 2009

### e(ho0n3

How can you modify this so that it works for $1 - x + x^2 - \cdots$

There's no need for the Comparison Test. Just note that

$$\sum_{n=1}^\infty \frac{1}{n + 1} = \frac{1}{2} + \frac{1}{3} + \cdots = \sum_{n=2}^\infty \frac{1}{n}$$

6. Apr 9, 2009

### latentcorpse

would it have a numerator of $1-(-1)^nx^{n+1}$?

7. Apr 9, 2009

### e(ho0n3

I don't know this of the top of my head, but if you write up what you did to get that, I can look it over.

8. Apr 9, 2009

### HallsofIvy

Staff Emeritus
WHAT would? You started talking about
$$\sum_{i= 0}^n (-x)^i[/itex] and then changed to [tex]\sum_{n=0}^\infty \frac{1}{n+1}$$

Are you back to the first question again?

9. Apr 10, 2009

### Chaos2009

Looking at your equation right here, I notice one problem that would fix it rather quickly. When you tried to find the nth partial sum, you are took:

$\sum_{k=1}^{n} (-x)^k$

The only problem with this is that when n = 1, you get -x just like you got. You accidentally dropped the first term when you switched to sum notation. Just change it to:

$\sum_{k=0}^{n} (-x)^k$

This way, 1 is the first term. Your new equation is:

$\sum_{k=0}^{n} a*r^k = \frac{a(r^{n + 1} - 1)}{r - 1}$