In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function’s derivatives at a single point.
The Taylor series of a real or complex-valued function \(ƒ(x)\) that is infinitely differentiable at a real or complex number \(a\) is the power series
\[f(x) = f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+ \cdots. \]
which can be written in the more compact sigma notation as
\[ \sum_{n=0} ^ {\infty} \frac {f^{(n)}(a)}{n!} \, (x-a)^{n}\] where \(n!\) denotes the factorial of n and \(ƒ^{(n)}(a)\) denotes the nth derivative of \(ƒ\) evaluated at the point a.
The derivative of order zero of \(ƒ\) is defined to be ƒ itself and \((x − a)^0\) and \(0!\) are both defined to be 1.
When a = 0, the series is also called a Maclaurin series.
Problem
Find the value of
\[\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \]
Solution
Using the Taylor Series
\[ln(x + 1) = x - \frac{1}{2} x^2 + \frac{1}{2} x^3 - \frac{1}{4} x^4 + ... = \sum^{\infty}_{n=1} \frac{x^n}{n} (-1)^{n-1}\]
at \(x = 1\),
\[\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + = ln(1 + 1) = ln(2)\]