Leibniz Series

The partial sums overshoot and undershoot π4\frac{\pi}{4} in turn, closing in from both sides. The gap between two neighboring partial sums is exactly the error bound.

π4=1−13+15−17+⋯\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots
∣SN−π4∣≤12N+1\left|S_N-\frac{\pi}{4}\right|\le\frac{1}{2N+1}

Examples & Exercises

Prove the alternating-series error bound ∣SN−π4∣≤12N+1\left|S_N - \frac{\pi}{4}\right| \le \frac{1}{2N+1}.

∣L−SN∣≤∣SN+1−SN∣=12N+1|L - S_N| \le |S_{N+1} - S_N| = \frac{1}{2N+1}
N=12N = 12: error ≤1/25=0.04\le 1/25 = 0.04 — squeezed from both sides.
Show proof
  1. For an alternating series ∑k=0∞(−1)kak\sum_{k=0}^{\infty}(-1)^k a_k with ak↓0a_k \downarrow 0, the partial sums oscillate inward: even ones increase, odd ones decrease, and the limit LL always lies between two consecutive partial sums.
  2. Hence ∣L−SN∣≤∣SN+1−SN∣=aN+1|L - S_N| \le |S_{N+1} - S_N| = a_{N+1}. For the Leibniz series aN+1=12N+1a_{N+1} = \frac{1}{2N+1}, so ∣SN−π4∣≤12N+1\left|S_N - \frac{\pi}{4}\right| \le \frac{1}{2N+1}.
  3. (Convergence of the series itself follows from the alternating series test, since ak=12k+1↓0a_k = \frac{1}{2k+1} \downarrow 0.)

How many terms are needed for three correct decimal places? Use the error bound.

12N+1<0.0005  ⟺  N≥1000\frac{1}{2N+1} < 0.0005 \iff N \ge 1000
Even at N=60N = 60 only two decimals — very slow convergence.
Show proof
  1. Three correct decimals (rounding) requires error <0.0005< 0.0005.
  2. The error bound gives 12N+1<0.0005\frac{1}{2N+1} < 0.0005, i.e. 2N+1>20002N + 1 > 2000. Since 11999≈0.00050025>0.0005\frac{1}{1999} \approx 0.00050025 > 0.0005 at N=999N = 999, we need N≥1000N \ge 1000.
  3. So a thousand terms are needed — the notorious slowness of the Leibniz series.

Derive arctan⁡x=∑k=0∞(−1)kx2k+12k+1\arctan x = \sum_{k=0}^{\infty}\frac{(-1)^k x^{2k+1}}{2k+1} for ∣x∣<1|x| < 1, giving the Leibniz series at x=1x = 1. (The final step uses Abel's theorem, taken as known.)

arctan⁡x=∫0xdt1+t2=∑k=0∞(−1)kx2k+12k+1\arctan x = \int_0^x \frac{dt}{1+t^2} = \sum_{k=0}^{\infty}\frac{(-1)^k x^{2k+1}}{2k+1}
Where the series comes from: the power series of arctan⁡\arctan at x=1x = 1.
Show proof
  1. For ∣t∣<1|t| < 1, 11+t2=∑k=0∞(−t2)k\frac{1}{1+t^2} = \sum_{k=0}^{\infty}(-t^2)^k (geometric series).
  2. On [0,x][0, x] with ∣x∣<1|x| < 1 the convergence is uniform, so termwise integration is legitimate: arctan⁡x=∫0x∑k=0∞(−1)kt2k dt=∑k=0∞(−1)kx2k+12k+1\arctan x = \int_0^x \sum_{k=0}^{\infty}(-1)^k t^{2k}\,dt = \sum_{k=0}^{\infty}\frac{(-1)^k x^{2k+1}}{2k+1}.
  3. Substituting x=1x = 1 involves a conditionally convergent series, so Abel's theorem is needed — taken as known, it gives arctan⁡1=π4\arctan 1 = \frac{\pi}{4} as the Leibniz series.

6. Series