Find the Taylor series of (1 - x)^k about x = 0, where k is any real number. Give the general term and the radius of convergence.

Step-by-step solution

1. 1

   Use the generalised binomial theorem rather than derivatives. In principle the Maclaurin series is $\sum\frac{f^{(n)}(0)}{n!}x^{n}$, but computing $f^{(n)}(x)=(-1)^{n}k(k-1)\cdots(k-n+1)(1-x)^{k-n}$ by hand is laborious. The binomial series packages the same result:

   $(1+t)^{k}=\sum_{n=0}^{\infty}\binom{k}{n}t^{n},\qquad \binom{k}{n}=\frac{k(k-1)\cdots(k-n+1)}{n!},$

   valid for $|t|<1$ and any real $k$ — not just non-negative integers.

2. 2

   Substitute $t=-x$. This is the only adjustment needed:

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

   The factor $(-1)^{n}$ is where every sign in the expansion comes from.

3. 3

   Write out the first few terms. Expanding the coefficients:

   $(1-x)^{k}=1-kx+\frac{k(k-1)}{2!}x^{2}-\frac{k(k-1)(k-2)}{3!}x^{3}+\cdots$

   Check against a familiar case: with $k=-1$ every coefficient $(-1)^{n}\binom{-1}{n}$ equals $1$, recovering the geometric series $\frac{1}{1-x}=1+x+x^{2}+\cdots$ ✓.

4. 4

   Note when the series terminates. If $k$ is a non-negative integer, the factor $(k-n)$ makes $\binom{k}{n}=0$ for all $n>k$, so the sum stops and reduces to the ordinary finite binomial expansion — for instance $(1-x)^{3}=1-3x+3x^{2}-x^{3}$. For every other $k$ the series is genuinely infinite.

5. 5

   State the radius of convergence and verify numerically. By the ratio test, $\left|\frac{a_{n+1}}{a_{n}}\right|=\left|\frac{k-n}{n+1}\right||x|\to|x|$, so the series converges for

   $|x|<1$

   and diverges for $|x|>1$; endpoint behaviour depends on $k$ (it converges at both ends when $k>0$). Numeric check with $k=2.5$ and $x=0.3$: summing sixty terms gives $0.409963413001697$ against $(0.7)^{2.5}=0.409963413001697$ ✓.