Building a new power series without differentiating from scratch
Substitute into, combine, differentiate, or integrate a known series — almost never start over.
Before computing a single derivative, check whether the target function is a disguised version of something already in the standard library (geometric, e^x, \sin x, \cos x, \ln(1+x), or a binomial power). Substituting a new expression into a known series' variable, multiplying by x^k, or differentiating/integrating term by term is almost always faster — and less error-prone — than computing \frac{f^{(n)}(a)}{n!} directly for a high-order derivative.
The standard library
Six series, each centered at 0, that every other series in this chapter is built from by substitution, combination, or term-by-term calculus: e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}, \qquad (-\infty < x < \infty) \sin x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}, \qquad (-\infty < x < \infty) \cos x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}, \qquad (-\infty < x < \infty) \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n, \qquad (-1 < x < 1) \ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}, \qquad (-1 < x \le 1) \arctan x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}, \qquad (-1 \le x \le 1)
For |x| < 1, \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots, with radius of convergence R=1. Replacing x by any expression u(x) gives a power series for \dfrac{1}{1-u(x)} wherever |u(x)|<1, and multiplying both sides by a fixed function of x produces a power series for that product.
For |r| < 1, S = \sum_{k=1}^{\infty} a_1 r^{\,k-1} = \frac{a_1}{1 - r}. The series diverges (has no finite sum) when |r| \ge 1.
For any real number k and |x|<1, (1+x)^k = \sum_{n=0}^{\infty} \binom{k}{n} x^n, \qquad \binom{k}{n} = \frac{k(k-1)(k-2)\cdots(k-n+1)}{n!}, \quad \binom{k}{0}=1. This is the Maclaurin series of (1+x)^k. When k is a nonnegative integer, \binom{k}{n}=0 for n>k and the series terminates, recovering the ordinary Binomial Theorem for integer powers; for any other k (negative, or non-integer) the series never terminates and has radius of convergence R=1.
Combining and transforming
If \displaystyle f(x) = \sum_{n=0}^{\infty} a_n(x-c)^n has radius of convergence R>0, then for every x with |x-c|<R:
f'(x) = \sum_{n=1}^{\infty} n\,a_n(x-c)^{n-1}, \qquad \int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{a_n}{n+1}(x-c)^{n+1}.Both new series have the SAME radius of convergence R as the original (though the behavior at the two endpoints can change). Differentiating or integrating a power series is legal term by term anywhere strictly inside the interval of convergence. No extra hypothesis is required beyond convergence itself.
If \displaystyle\sum_{n=1}^{\infty} a_n = A and \displaystyle\sum_{n=1}^{\infty} b_n = B both converge, and c is a constant, then:
- \displaystyle\sum_{n=1}^{\infty} (a_n \pm b_n) = A \pm B
- \displaystyle\sum_{n=1}^{\infty} c \cdot a_n = c A
A convergent series plus a DIVERGENT one always diverges. Two divergent series added together can do anything, including converge.
Show proof
Each partial sum of the combined series splits as \sum_{k=1}^n (a_k \pm b_k) = \sum_{k=1}^n a_k \pm \sum_{k=1}^n b_k, a finite sum being rearranged, which is always legal. Taking n \to \infty on both sides and applying the sequence limit laws to the two partial-sum sequences (which converge to A and B by hypothesis) gives the result directly. The constant-multiple case is the same argument with the constant-multiple limit law instead of the sum law.