Geometric Progression: Calculating the nth Partial Sum

[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?

 

[e(ho0n3]

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

 

[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}

 

[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?

 

[e(ho0n3]

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}

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

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?

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}

 

[latentcorpse]

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

 

[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.

 

[HallsofIvy]

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

WHAT would? You started talking about
\sum_{i= 0}^n (-x)^i[/itex] <br /> and then changed to <br /> \sum_{n=0}^\infty \frac{1}{n+1}<br /> <br /> Are you back to the first question again?

 

[Chaos2009]

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?

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}