Adding Series, What is wrong here?

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Homework Statement


With binomial expansions,
[itex]\frac{x}{1-x}[/itex] = [itex]\sum[/itex][itex]^{n=\infty}_{n=1}[/itex]xn
[itex]\frac{x}{x-1}[/itex] = [itex]\sum[/itex][itex]^{n=\infty}_{n=0}[/itex]x-n

Adding these series yields:

[itex]\sum[/itex][itex]^{n=\infty}_{n=-\infty}[/itex]xn=0
This is nonsense, but what went wrong here?

The Attempt at a Solution


Obviously, [itex]\frac{x}{1-x}[/itex]+[itex]\frac{x}{x-1}[/itex]=0 (1)

It's clear that they tried to transform [itex]\frac{x}{x-1}[/itex] = [itex]\sum[/itex][itex]^{n=\infty}_{n=0}[/itex]x-n into [itex]\frac{x}{x-1}[/itex] = [itex]\sum[/itex][itex]^{n=0}_{n=-\infty}[/itex]xn and then substitute into equation (1) and get an answer of zero.

From there, I'm not sure how in the world they manipulated that series to get a neg. infinity to show up in the limits, and where the mistake in that is.
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement


With binomial expansions,
[itex]\frac{x}{1-x}[/itex] = [itex]\sum[/itex][itex]^{n=\infty}_{n=1}[/itex]xn
[itex]\frac{x}{x-1}[/itex] = [itex]\sum[/itex][itex]^{n=\infty}_{n=0}[/itex]x-n
The second is wrong. x- 1= -(1- x) so the second series is just
[itex]\sum_{n=1}^\infty -x^n[/itex], not [itex]\sum_{n=1}^\infty x^{-n}[/itex].

Adding these series yields:

[itex]\sum[/itex][itex]^{n=\infty}_{n=-\infty}[/itex]xn=0
This is nonsense, but what went wrong here?

The Attempt at a Solution


Obviously, [itex]\frac{x}{1-x}[/itex]+[itex]\frac{x}{x-1}[/itex]=0 (1)

It's clear that they tried to transform [itex]\frac{x}{x-1}[/itex] = [itex]\sum[/itex][itex]^{n=\infty}_{n=0}[/itex]x-n into [itex]\frac{x}{x-1}[/itex] = [itex]\sum[/itex][itex]^{n=0}_{n=-\infty}[/itex]xn and then substitute into equation (1) and get an answer of zero.

From there, I'm not sure how in the world they manipulated that series to get a neg. infinity to show up in the limits, and where the mistake in that is.
 
  • #3
jbunniii
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Homework Statement


With binomial expansions,
[itex]\frac{x}{1-x}[/itex] = [itex]\sum[/itex][itex]^{n=\infty}_{n=1}[/itex]xn
[itex]\frac{x}{x-1}[/itex] = [itex]\sum[/itex][itex]^{n=\infty}_{n=0}[/itex]x-n

Adding these series yields:

[itex]\sum[/itex][itex]^{n=\infty}_{n=-\infty}[/itex]xn=0
This is nonsense, but what went wrong here?
The first series converges only if |x| < 1. The second converges only if |x| > 1. Therefore there is no x for which you can add the series and get a meaningful result.
 
  • #4
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The second is wrong. x- 1= -(1- x) so the second series is just
[itex]\sum_{n=1}^\infty -x^n[/itex], not [itex]\sum_{n=1}^\infty x^{-n}[/itex].

I don't understand this. I realize this myself, and if I were deriving the series myself, I'd use that technique and get the same result you. Also, when I use Maple to get the series, the result also matches yours. But, reference tables I have on hand show series matching what was printed in the homework.
 
  • #5
jbunniii
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If |x| > 1, then
[tex]\sum_{n = 0}^{\infty} x^{-n} = \frac{1}{1 - x^{-1}} = \frac{x}{x - 1}[/tex]
so the series expansion is correct. The problem is that the region of convergence is incompatible with that of the first series.
 
  • #6
Ray Vickson
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I don't understand this. I realize this myself, and if I were deriving the series myself, I'd use that technique and get the same result you. Also, when I use Maple to get the series, the result also matches yours. But, reference tables I have on hand show series matching what was printed in the homework.

If you expand x/(1-x) as a series in t = 1/x you will get the printed result. One expansion applies for small |x| and the other applies for large |x|.

RGV
 

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