Find the domain of convergence and the sum of the series \[\sum_{n\ge0}(-1)^n\frac{x^{2n+1}}{2n+1}.\]
Show how one may use the sum of the series to provide an approximation for \(\pi\) up to three decimals. Be sure to provide all technical details.
Solution
Goal: Find the domain of convergence and the sum of \(\sum_{n\ge 0}(-1)^n\frac{x^{2n+1}}{2n+1}\), and use it to approximate \(\pi\) to three decimals.
1. The series is the Taylor series of \(\arctan x\) at \(0\): \(\arctan x = \sum_{n\ge0}(-1)^n\frac{x^{2n+1}}{2n+1}\).Proof: \(\arctan' x = \frac{1}{1+x^2} = \sum_{n\ge0}(-1)^n x^{2n}\) (geometric series, \(|x| < 1\)), and termwise integration from \(0\) to \(x\) gives the claim.
2. Domain of convergence: \([-1, 1]\).
2.1. For \(|x| < 1\): absolute convergence.Proof: \(\left|(-1)^n\frac{x^{2n+1}}{2n+1}\right| \le |x|^{2n+1}\), geometric with ratio \(|x|^2 < 1\).
2.2. At \(x = 1\): the alternating series \(\sum (-1)^n \frac{1}{2n+1}\) converges.Proof: Leibniz test: \(\frac{1}{2n+1} \downarrow 0\).
2.3. At \(x = -1\): \(\sum (-1)^n \frac{(-1)^{2n+1}}{2n+1} = \sum (-1)^{n+1}\frac{1}{2n+1}\), also an alternating series with terms decreasing to \(0\): converges.Proof: same Leibniz test.
2.4. For \(|x| > 1\): the terms do not tend to \(0\).Proof: \(\left|(-1)^n\frac{x^{2n+1}}{2n+1}\right| = \frac{|x|^{2n+1}}{2n+1} \to \infty\) (exponential beats linear), so the series diverges.
2.5. Q.E.D. Proof: <2>1–<2>4 show the domain is \([-1, 1]\).
3. On \([-1,1]\) the sum equals \(\arctan x\) (continuous extension).Proof: the series converges uniformly on \([-r, r]\) for \(r < 1\) and pointwise on \([-1,1]\) (by <1>2); Abel’s theorem / continuity of the sum gives equality on the closed interval.At \(x = 1\): \(\arctan 1 = \frac{\pi}{4}\).
4. Approximation of \(\pi\): \(\pi = 4\arctan 1 = 4\sum_{n\ge0}\frac{(-1)^n}{2n+1}\).
4.1. The alternating series at \(x = 1\) has error bounded by the first omitted term: \(\left|\frac{\pi}{4} - \sum_{n=0}^{N}\frac{(-1)^n}{2n+1}\right| \le \frac{1}{2N+3}\).Proof: alternating-series remainder bound.
4.2. To get \(\pi\) to three decimals (error \(< 0.0005\), i.e. \(\pi/4\) to error \(< 0.000125\)), need \(\frac{1}{2N+3} < 1.25\times10^{-4}\), i.e. \(2N + 3 > 8000\), \(N \ge 4000\).Proof: solve the bound of 1: error in \(\pi\) is \(4/(2N+3) < 0.0005 \Leftrightarrow 2N + 3 > 8000\).
4.3. Compute: the partial sum through \(n=4000\) (that is, \(4001\) terms) gives \(\pi \approx 3.141\ldots\); e.g. the partial sums oscillate around \(\pi/4 \approx 0.785398\), and \(\pi \approx 3.14159\).Proof: the alternating-series bound of 1 guarantees three-decimal accuracy: \(|\pi - 4S_N| \le 4/(2N+3) < 0.0005\).(This is the Leibniz formula for \(\pi\); it converges slowly — \(\sim 4000\) terms — but the bound is rigorous.)
4.4. Q.E.D. Proof: <2>1–<2>3 give the approximation procedure with all details.