Knowing that a series converges is only half the story; you usually want its value. A computer adds up $S_N$, but how big is the leftover tail? The The Integral Test and p-Series and the alternating series test both come with error bounds.
Why it works
Write $S = S_N + R_N$, where the remainder $R_N = a_{N+1} + a_{N+2} + \cdots$ is a tail of the series. For $a_n = f(n)$ with $f$ positive and decreasing, the same rectangle picture as in the integral test traps the tail between two integrals:
$$\int_{N+1}^{\infty} f(x)\,dx \le R_N \le \int_{N}^{\infty} f(x)\,dx.$$
For an alternating series with decreasing terms, $S$ always sits between consecutive partial sums, so $|R_N| \le b_{N+1}$. For a tail you can compare with a geometric series, use the geometric sum.
Key idea
Integral bound: $\int_{N+1}^{\infty} f \le R_N \le \int_N^{\infty} f$. Alternating bound: $|R_N| \le b_{N+1}$. Comparison: if $0 \le a_n \le b_n$, then $R_N \le \sum_{n=N+1}^{\infty} b_n$.
Worked example: Integral bounds
Estimate $\sum_{n=1}^{\infty}\frac{1}{n^{3}}$ using $S_{10}$, and bound the error.
- $\int_M^{\infty}\frac{dx}{x^{3}} = \frac{1}{2M^{2}}$. So $\frac{1}{2\cdot 11^{2}} \le R_{10} \le \frac{1}{2\cdot 10^{2}}$, that is $0.00413 \le R_{10} \le 0.005$.
- $S_{10} \approx 1.19753$, so $1.20166 \le S \le 1.20253$.
- Taking the middle, $S \approx 1.2021$; the true value is about $1.2020569$.
Answer: $S \approx 1.202$, with error under $0.0005$ using the midpoint
Worked example: An alternating bound
How close is $S_5$ to $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}$?
- The terms $\frac{1}{n!}$ decrease to 0, so $|S - S_N| \le b_{N+1}$.
- $S_5$ uses $n = 0$ to $4$, so the first omitted term is $\frac{1}{5!} = \frac{1}{120} \approx 0.0083$.
- (The sum is $\frac{1}{e} \approx 0.3679$; $S_5 = 0.375$.)
Answer: Within $\frac{1}{120}$
Worked example: How many terms?
How many terms of $\sum\frac{1}{n^{2}}$ guarantee an error of at most $0.001$?
- $R_N \le \int_N^{\infty}\frac{dx}{x^{2}} = \frac{1}{N}$.
- $\frac{1}{N} \le 0.001$ when $N \ge 1000$.
- So 1000 terms are enough. (Slowly converging series need many terms; this is why better methods matter.)
Answer: $N = 1000$
Common mistake
Using the alternating bound on a series that is not alternating, or off by one. For $\sum\frac{1}{n^{2}}$ the error after $N$ terms is not $a_{N+1}$; the tail is many terms, about $\frac{1}{N}$ in total. And for alternating series the bound is the first term you left out, $b_{N+1}$, not the last one you used. Check which terms $S_N$ contains before writing the bound.