Leibniz Series | Real Analysis

The partial sums overshoot the target, then undershoot it, closing in from both sides. It is a sandwich that keeps tightening.

Add one term and you jump past the target; the next term pulls you back. The back-and-forth swing shrinks every time. So the target always sits between two neighboring partial sums. That gap is exactly the error bound.

For example, a ⚽ ball rolls past a line, comes back, and crosses it again. Each crossing overshoots by less than the one before. The ball has no choice but to settle on the line.

Examples & Exercises

With $S_N := \sum_{k=0}^{N} \frac{(-1)^k}{2k+1}$ , prove the alternating-series error bound $\left|S_N - \frac{\pi}{4}\right| \le \frac{1}{2N+3}$ .

|L - S_N| \le |S_{N+1} - S_N| = a_{N+1} = \frac{1}{2N+3}, \qquad S_N = \sum_{k=0}^{N} \frac{(-1)^k}{2k+1}

N = 12

(13 terms): error

\le 1/27 \approx 0.037

— squeezed from both sides.

Show proof

Put $a_k = \frac{1}{2k+1} \downarrow 0$ and $S_N = \sum_{k=0}^{N} (-1)^k a_k$ .
Prove the nesting by paired differences: $S_{2m+2} - S_{2m} = -a_{2m+1} + a_{2m+2} = -(a_{2m+1} - a_{2m+2}) \le 0$ , so the even partial sums $(S_{2m})$ decrease; and for $m \ge 1$ , $S_{2m+1} - S_{2m-1} = a_{2m} - a_{2m+1} \ge 0$ , so the odd partial sums $(S_{2m+1})$ increase.
Also $S_{2m} - S_{2m+1} = a_{2m+1} > 0$ .
The sequence $(S_{2m})$ is decreasing and bounded below by the odd sums, and $(S_{2m+1})$ is increasing and bounded above, so both converge.
The gap $S_{2m} - S_{2m+1} = a_{2m+1} \to 0$ , so both limits equal a common $L$ , with $S_{2m+1} \le L \le S_{2m}$ for every $m$ .
Hence $L$ always lies between two consecutive partial sums, and $|L - S_N| \le |S_{N+1} - S_N| = a_{N+1}$ .
Here $a_{N+1} = \frac{1}{2(N+1)+1} = \frac{1}{2N+3}$ .
Identify the limit: this series is the arctan series of the problem below at $x = 1$ , so by Abel continuity its sum is $\arctan 1 = \frac{\pi}{4}$ . That is, $L = \frac{\pi}{4}$ , hence $\left|S_N - \frac{\pi}{4}\right| \le \frac{1}{2N+3}$ .

How many terms suffice to guarantee three correct decimal places?

Use $S_N = \sum_{k=0}^{N} \frac{(-1)^k}{2k+1}$ from the previous problem and its error bound to give a sufficient number.

\frac{1}{2N+3} < 0.0005 \iff N \ge 999 \quad \text{(1000 terms suffice)}

Even with 60 terms only two decimals — very slow convergence.

Show proof

Three correct decimals (rounding) requires error $< 0.0005$ .
The error bound gives the sufficient condition $\frac{1}{2N+3} < 0.0005$ , i.e. $2N + 3 > 2000$ , $N > 998.5$ .
So $N \ge 999$ suffices.
Check: at $N = 999$ , $\frac{1}{2001} \approx 0.00049975 < 0.0005$ ; at $N = 998$ the bound is $\frac{1}{1999} \approx 0.00050025 > 0.0005$ , so this bound does not guarantee the accuracy there (that does not mean the actual error is larger).
So $N = 999$ , i.e. 1000 terms ( $k = 0, \dots, 999$ ), suffice to guarantee the accuracy — this is a sufficient guarantee, not a claim of minimality. The notorious slowness of the Leibniz series.

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

\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

x = 1

Show proof

For $|t| < 1$ , $\frac{1}{1+t^2} = \sum_{k=0}^{\infty}(-t^2)^k$ (geometric series).
For fixed $|x| < 1$ , every $t$ between $0$ and $x$ satisfies $\lvert (-t^2)^k \rvert = \lvert t \rvert^{2k} \le \lvert x \rvert^{2k}$ , and $\sum_k \lvert x \rvert^{2k}$ is a convergent geometric series, so the Weierstrass M-test gives uniform convergence on that interval.
Hence termwise integration is legitimate: for $0 \le x < 1$ , $\arctan x = \int_0^x \sum_{k=0}^{\infty}(-1)^k t^{2k}\,dt = \sum_{k=0}^{\infty}\frac{(-1)^k x^{2k+1}}{2k+1}$ .
For $x < 0$ the interval flips: the same bound holds on $[x, 0]$ , $\int_0^x = -\int_x^0$ , and both sides are odd functions, so the same formula holds.
Substituting $x = 1$ involves a conditionally convergent series, so Abel's theorem is needed — taken as known, it gives $\arctan 1 = \frac{\pi}{4}$ as the Leibniz series.