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Is there a way to derive the Maclaurin series for $\frac{1}{(1-x)}$ after finding the Maclaurin series for $(1+x)^n$ which is $\displaystyle\sum\limits_{k=0}^\infty \frac{f^k(0)}{k!}*x^k$.

From the original equation I could substitute $-1$ for $n$ and $-x$ for $x$ but I don't see how I could do that in the summation. The intention of this is getting the next series without having to do the expansion from scratch.

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In my opinion, neither of the two responses given thus far directly addresses the question. For that matter, I don't see how to answer it either, and I really don't see what the point of the question is: there is no lack of good ways to derive the Maclaurin series for $\frac{1}{1-x}$. My sympathies to the questioner. – Pete L. Clark Sep 15 '10 at 5:45
Pete: I was hoping he'd be able to figure out how to derive the binomial series first, and then make the appropriate substitutions. – J. M. Sep 15 '10 at 9:55
@J.M.: What you're giving is a common generalization of the two series. As I understand the problem, that doesn't answer it. Of course what you're doing is more general and more useful than what the question wants, but still... – Pete L. Clark Sep 15 '10 at 11:07
Looking at Andy's answer, I think I understand J.M.'s better: you two are both saying that the "$n$" in $(1+x)^n$ is not intended to be a positive integer, i.e., the question really is how to derive the Maclaurin series for $\frac{1}{1-x}$ from the binomial series. Mea culpa: I wasn't reading the question that way. Moral: don't use $n$ for a variable that is allowed to take real (or complex...) values. Some people will be confused! – Pete L. Clark Sep 15 '10 at 15:43
No worries Pete, I have to confess I had to overcome my usual convention that $n$ is a variable reserved for nonnegative integers. It was pretty hard, too. :) – J. M. Sep 15 '10 at 15:50

The binomial series works for any real $n$.

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It works for complex $n$; in fact it works for formal $n$. – Qiaochu Yuan Sep 15 '10 at 3:10
Qiaochu: I know (for the uninitiated: because the gamma function is finite except at nonpositive integers, and the series for positive $n$ terminates because all higher terms have the form $1/\infty$), but I didn't want to distract him with that. :) – J. M. Sep 15 '10 at 3:17

Elaborating on J.M. answer, you can procede this way:

$$(1+x)^n = 1+ n x + \frac{1}{2}(n-1) n x^2 + \frac{1}{6}(n-2)(n-1)n x^3 + \ldots$$, more generally, $$(1+x)^\alpha = 1 + \sum_{k = 1}^{n} \binom{\alpha}{k} x^k + o(x^n)$$ for any real $\alpha$, where $\binom{\alpha}{k} = \frac{(\alpha -1)(\alpha-2)\ldots (\alpha - k +1)}{k!}$ is the binomial coefficient extended to any real number. Then, you get that $$(1-x)^n = 1 - nx + \frac{1}{2}(n-1)n x^2 - \frac{1}{6}(n-2)(n-1)n x^3 + \ldots$$

then, you only have to plug in $n = -1$ to get $$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots$$

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The expression $\displaystyle \sum_{k=0}^{\infty} \frac{f^k(0)}{k!} x^k$ is not just the Maclaurin series for functions of the form $f(x)=(1+x)^n$; it is a formula for the Maclaurin series of any function infinitely differentiable at $0$. It gives us the power series in $x$ with all derivatives at $x=0$ matching the derivatives of $f(x)$ at $x=0$.

If you want to use this formula to find the Maclaurin series for $f(x)=\frac{1}{1-x}$, you need to compute all of the derivatives of $f(x)$. --Fortunately, there is a simple pattern you can observe and prove (by induction).

Once you know the derivatives, you can plug in $x=0$, simplify your summation formula, and you'll have your answer.

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Try division?

$\dfrac {1}{1-x} = 1 + x + x^2 + x^3 ...$

(edit: I'm not sure why you would try to derive it from the series for $(1+x)^n$ )

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It was a question in my book which I couldn't understand. I know how to find the series through expansion. – Planeman Sep 15 '10 at 2:15

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