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Geometric Progression

  • #1
1,444
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what is the nth partial sum of

[itex]1-x+x^2-x^3+..[/itex]

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

i have [itex]\sum_{k=1}^{n} ar^k=\frac{a(r^{n+1}-r)}{r-1}[/itex]

ok but then when i sub in i get messed up i get [itex]\frac{(-x)^{n+1}+x}{-(1+x)}[/itex] 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???
 

Answers and Replies

  • #2
1,357
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Do you know how to derive the nth partial sum of 1 + x + x^2 + ... ?
 
  • #3
1,444
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off the top of my head isn't it something like

[itex](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}[/itex]
 
  • #4
1,444
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also, another thing i'm having trouble seeing is why [itex]\sum \frac{1}{n+1}[/itex] diverges.
my initial instinct was to use the copmarison test with [itex]\sum \frac{1}{n}[/itex] as this diverges.

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

what's going on here?
 
  • #5
1,357
0
off the top of my head isn't it something like

[itex](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}[/itex]
How can you modify this so that it works for [itex]1 - x + x^2 - \cdots[/itex]

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

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

what's going on here?
There's no need for the Comparison Test. Just note that

[tex]\sum_{n=1}^\infty \frac{1}{n + 1} = \frac{1}{2} + \frac{1}{3} + \cdots = \sum_{n=2}^\infty \frac{1}{n}[/tex]
 
  • #6
1,444
0
would it have a numerator of [itex]1-(-1)^nx^{n+1}[/itex]?
 
  • #7
1,357
0
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
HallsofIvy
Science Advisor
Homework Helper
41,833
955
would it have a numerator of [itex]1-(-1)^nx^{n+1}[/itex]?
WHAT would? You started talking about
[tex]\sum_{i= 0}^n (-x)^i[/itex]
and then changed to
[tex]\sum_{n=0}^\infty \frac{1}{n+1}[/tex]

Are you back to the first question again?
 
  • #9
81
0
what is the nth partial sum of

[itex]1-x+x^2-x^3+..[/itex]

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

i have [itex]\sum_{k=1}^{n} ar^k=\frac{a(r^{n+1}-r)}{r-1}[/itex]

ok but then when i sub in i get messed up i get [itex]\frac{(-x)^{n+1}+x}{-(1+x)}[/itex] 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:

[itex]\sum_{k=1}^{n} (-x)^k[/itex]

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:

[itex]\sum_{k=0}^{n} (-x)^k[/itex]

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

[itex]\sum_{k=0}^{n} a*r^k = \frac{a(r^{n + 1} - 1)}{r - 1}[/itex]
 

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