Question

Find a power series representation for the function and determine the interval of convergence. See first post for the problem, second post will be work i have so far.

Answer 1

\[f(x)=\frac{ 1 }{ (1+x)^{2} } \]
(Hint: use the integral of f(x).)

now lets see if i am doing this right (next post)

Answer 2

Stuck?

Answer 3

use this put m=2

Answer 4

sorry put m=-2

Answer 5

\[\large \int\limits \frac{ 1 }{ (1+x)^{2} }\]
then i substituted
u=1+x
du=dx

\[\large \int\limits u ^{-2}\]
\[\large \frac{ -1 }{ 1+x }=\sum_{n=0}^{\infty}(-x)^{n}\]
so the interval would be:
-1<x<1

Answer 6

Your series is off by a bit, I think:
\[-\frac{1}{1+x}=-\frac{1}{1-(-x)}=-\sum_{n=0}^{\infty}(-x)^n\]

Answer 7

So now you have the series for the integral of \(f(x)\). Differentiate this series and you get the series for \(f(x)\).

Answer 8

\[\large - \frac{ d }{dx } \sum_{n=0}^{\infty} (-x)^{n}= -\sum_{n=0}^{\infty} (-nx)^{n-1}\]
so the interval would still be:
\[\left| -x \right|<1\]
\[-1<x<1\]
or no?

Answer 9

I've since forgotten the summation differentiation/integraion rules, so I'll work out the series term-by-term. You'd get the same result.
\[-\sum_{n=0}^\infty(-x)^n=-\left(1-x+x^2-x^3+x^4-\cdots\right)\\
\frac{d}{dx}\left[-\sum_{n=0}^\infty(-x)^n\right]=-\left(-1+2x-3x^2+4x^3-5x^4+\cdots\right)\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~=1-2x+3x^2-4x^3+5x^4-\cdots\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\sum_{n=1}^\infty n(-x)^{n-1}\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\sum_{n=0}^\infty (n+1)(-x)^{n}\]
As you should be able to see, \(n\) does not take on the exponent.

But yes, your interval of convergence is still the same. The ratio test tells us that the series converges if \(\displaystyle\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}<1\), so you have
\[\lim_{n\to\infty}\frac{|(n+2)(-x)^{n+1}|}{|(n+1)(-x)^n|}=|-x|=|x|\]
which says the series converges if \(|x|<1\), i.e. \(-1<x<1\).