Adding Series, What is wrong here?

[jtleafs33]

Homework Statement

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

Adding these series yields:

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

The Attempt at a Solution

Obviously, \frac{x}{1-x}+\frac{x}{x-1}=0 (1)

It's clear that they tried to transform \frac{x}{x-1} = \sum^{n=\infty}_{n=0}x^-n into \frac{x}{x-1} = \sum^{n=0}_{n=-\infty}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.

 

[HallsofIvy]

Homework Statement

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

The second is wrong. x- 1= -(1- x) so the second series is just
\sum_{n=1}^\infty -x^n, not \sum_{n=1}^\infty x^{-n}.

Adding these series yields:

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

The Attempt at a Solution

Obviously, \frac{x}{1-x}+\frac{x}{x-1}=0 (1)

It's clear that they tried to transform \frac{x}{x-1} = \sum^{n=\infty}_{n=0}x^-n into \frac{x}{x-1} = \sum^{n=0}_{n=-\infty}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.

 

[jbunniii]

Homework Statement

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

Adding these series yields:

\sum^{n=\infty}_{n=-\infty}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.

 

[jtleafs33]

The second is wrong. x- 1= -(1- x) so the second series is just
\sum_{n=1}^\infty -x^n, not \sum_{n=1}^\infty x^{-n}.

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.

 

[jbunniii]

If |x| > 1, then
\sum_{n = 0}^{\infty} x^{-n} = \frac{1}{1 - x^{-1}} = \frac{x}{x - 1}
so the series expansion is correct. The problem is that the region of convergence is incompatible with that of the first series.

 

[Ray Vickson]

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