Estimating the Value of a Series

Most convergent series can't be summed exactly, so we add the first $n$ terms and ask how big the remainder is:

$$S = s_n + R_n, \qquad R_n = a_{n+1} + a_{n+2} + \cdots$$

1. Integral test remainders

If $a_k = f(k)$ with $f$ positive, continuous and decreasing (10.08):

$$\int_{n+1}^{\infty} f(x)\,dx \;\le\; R_n \;\le\; \int_n^{\infty} f(x)\,dx$$

Adding $s_n$ traps the sum in an interval:

$$s_n + \int_{n+1}^{\infty} f \;\le\; S \;\le\; s_n + \int_n^{\infty} f$$

Taking the midpoint of that interval is usually a much better estimate than $s_n$ alone. For $\sum \frac{1}{k^2}$ with $n = 10$, the interval is $s_{10} + [\frac{1}{11}, \frac{1}{10}]$, and its midpoint, $1.6452$, is within $0.0003$ of $\frac{\pi^2}{6} = 1.6449$. The partial sum $s_{10} = 1.5498$ is off by almost $0.1$.

2. Alternating series remainders

If the series alternates and $b_k$ decreases to $0$ (10.10):

$$\lvert R_n\rvert \le b_{n+1}$$

the first term left out, and $R_n$ has the same sign as that term.

3. Ratio test remainders

If each term is at most $r$ times the one before, for some $r < 1$ (from $a_{n+1}$ on), the tail is smaller than a geometric series:

$$R_n \le a_{n+1}\left(1 + r + r^2 + \cdots\right) = \frac{a_{n+1}}{1 - r}$$

Use the largest ratio from that point on. If the ratios $\frac{a_{k+1}}{a_k}$ are decreasing, that's the first one, $\frac{a_{n+2}}{a_{n+1}}$. If they increase toward a limit $L < 1$, use $r = L$.

Example. For $e = \sum_{k=0}^{\infty} \frac{1}{k!}$, the ratios are $\frac{1}{k + 1}$, which decrease. Stopping after $\frac{1}{4!}$, the next term is $\frac{1}{5!}$ and the next ratio is $\frac{1}{6}$, so

$$R \le \frac{1/120}{1 - 1/6} = \frac{1}{100}$$

Which one?

Series	Bound
positive, decreasing, integrable terms	integrals
alternating, decreasing to $0$	the next term
factorials or powers (where the ratio test works)	geometric tail