find the power series representation for$$f(x)=\left(\frac{x}{2-x}\right)^{3}$$ I have $$\frac{1}{(1+x)^{3}}=-\frac{d}{dx}\frac{2}{(1+x)^{2}}=\frac{1}{2}\sum(-1)^{n}(n+2)(n+1)x^{n}$$

then got this $$\frac{1}{8}\left(\frac{1}{1-\frac{x}{2}}\right)^{3} =\sum_{n=0}^{\infty}\frac{1}{2^{n+4}}(n+2)(n+1)x^{n+3}$$ However, when I use the computer to sum it, it doesn't give me back $$f(x)=\left(\frac{x}{2-x}\right)^{3}$$ instead, it gives me $$f(x)=-\frac{x^{3}}{(-2+x)^{3}}$$

  • 3
    $\begingroup$ $$\left(\frac{x}{2-x}\right)^3 = \frac{x^3}{(2-x)^3} = \frac{x^3}{(-1)^3(-2+x)^3} = -\frac{x^{3}}{(-2+x)^{3}}$$ $\endgroup$
    – user137731
    Nov 23, 2016 at 3:33

1 Answer 1


For this problem, the generalized binomial theorem is your friend.

It states that, for any real $a$, $(1+x)^a =\sum_{n=0}^{\infty} \binom{a}{n} x^n $ for $|x| < 1$.

In this, $\binom{a}{n} =\dfrac{a(a-1)...(a-n+1)}{n!} $.

When $a$ is negative integer, $a=-m$,

$\begin{array}\\ \binom{a}{n} &=\dfrac{a(a-1)...(a-n+1)}{n!}\\ &=\dfrac{\prod_{k=0}^{n-1}(a-k)}{n!}\\ &=\dfrac{\prod_{k=0}^{n-1}(-m-k)}{n!}\\ &=(-1)^n\dfrac{\prod_{k=0}^{n-1}(m+k)}{n!}\\ &=(-1)^n\dfrac{\prod_{k=0}^{n-1}(m+n-1-k)}{n!}\\ &=(-1)^n\dfrac{(m+n-1)!}{(m-1)!n!}\\ &=(-1)^n\binom{m+n-1}{m-1}\\ \end{array} $

Therefore $(1+x)^{-m} =\sum_{n=0}^{\infty} (-1)^n\binom{m+n-1}{m-1} x^n $.

If we put $-x$ for $x$, as your problem has, we get $(1-x)^{-m} =\sum_{n=0}^{\infty} (-1)^n\binom{m+n-1}{m-1} (-x)^n =\sum_{n=0}^{\infty} \binom{m+n-1}{m-1} x^n $.

In your case, starting as you have done, but using this formula, we get

$\begin{array}\\ f(x) &=\left(\dfrac{x}{2-x}\right)^{3}\\ &=\dfrac{x^3}{8}(1-x/2)^{-3}\\ &=\dfrac{x^3}{8}\sum_{n=0}^{\infty} \binom{3+n-1}{2} (x/2)^n\\ &=\dfrac{x^3}{8}\sum_{n=0}^{\infty} \binom{n+2}{2} (x/2)^n\\ \end{array} $

  • $\begingroup$ @ marty cohen, this is pretty neat, I tested with Mathematica and the result is the same as I would use a regular power series method. However, in the proof, I do get a little confused on the 5th line onward. $$(-1)^{n}\frac{\prod^{n-1}_{k=0}(m+n-1-k)}{n!}$$ how did this come from the previous step $$(-1)^{n}\frac{\prod^{n-1}_{k=0}(m+k)}{n!}$$ $\endgroup$
    – DSL
    Nov 23, 2016 at 17:12
  • $\begingroup$ Replace k by n-1-k. This reverses the order of the product. $\endgroup$ Nov 23, 2016 at 23:42

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