Estimates of Sums and Test Selection, calculus lesson

Knowing that a series converges is often only half the job. A calculator or a computer still has to stop after finitely many terms, and you want to know how far off the answer can be. The tests from the previous lessons come with error bounds, and those bounds tell you in advance how many terms are enough. The second half of the job is choosing the right test in the first place.

Key terms

Absolute error

Definition If a convergent series has sum $S$ and you approximate it by the partial sum $s_n$ , the absolute error is

\lvert S - s_n \rvert = \lvert R_n \rvert,

where the remainder is

R_n = \sum_{k=n+1}^{\infty} a_k.

More generally, the absolute error of any approximation $A$ to $S$ is $\lvert S - A \rvert$ .

In plain words The absolute error is how far your estimate is from the true value, without caring about the direction.

Example For $\sum \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.644934$ the partial sum $s_{10} \approx 1.549768$ has absolute error about $0.095166$ .

Integral bounds

Definition Suppose $f$ is continuous, positive and decreasing on $[n, \infty)$ , $a_k = f(k)$ for $k \ge n$ , and $\sum a_k$ converges with sum $S$ . Then

s_n + \int_{n+1}^{\infty} f(x)\,dx \le S \le s_n + \int_n^{\infty} f(x)\,dx.

Taking the midpoint of this interval as the estimate gives an absolute error of at most half its width, which is

\frac{1}{2}\int_n^{n+1} f(x)\,dx.

In plain words Two improper integrals fence in the sum. The middle of the fence is usually a much better estimate than the partial sum alone.

Example Take $\sum \frac{1}{n^3}$ with $n = 5$ . The partial sum is

s_5 = 1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \frac{1}{125} \approx 1.185662,

and the two integrals are

\int_6^{\infty} x^{-3}\,dx = \frac{1}{72}, \qquad \int_5^{\infty} x^{-3}\,dx = \frac{1}{50},

1.199551 \le S \le 1.205662.

The midpoint $1.202606$ is guaranteed to be within half the width, $0.003056$ , of $S$ . The true sum is about $1.202057$ , so the actual error is only about $0.00055$ .

Geometric tail bounds

Definition Suppose $a_k \ne 0$ for $k \ge n + 1$ and there is a number $r$ with $0 \le r < 1$ such that

\left\lvert \frac{a_{k+1}}{a_k} \right\rvert \le r \quad \text{for all } k \ge n + 1.

Then $\lvert a_{n+1+j} \rvert \le \lvert a_{n+1} \rvert r^j$ for every $j \ge 0$ , so the series converges absolutely and

\lvert R_n \rvert \le \sum_{j=0}^{\infty} \lvert a_{n+1} \rvert r^j = \frac{\lvert a_{n+1} \rvert}{1 - r}.

In plain words If from some point on each term is at most a fixed fraction $r$ of the one before, the tail is smaller than a geometric series you can add exactly. This is the error bound that goes with the ratio test.

Example Take $e = \sum_{k=0}^{\infty} \frac{1}{k!}$ and stop after the term $k = 6$ . For $k \ge 7$ the ratio is $\frac{1}{k + 1} \le \frac{1}{8}$ , so the tail after $\frac{1}{6!}$ satisfies

0 < e - \sum_{k=0}^{6} \frac{1}{k!} \le \frac{1/7!}{1 - 1/8} = \frac{8}{7 \cdot 5040} \approx 0.000227.

Indeed $\sum_{k=0}^{6} \frac{1}{k!} \approx 2.718056$ and $e \approx 2.718282$ , a difference of about $0.000226$ . The bound is almost exactly the true error.

Alternating error estimates

Definition If $\sum (-1)^{k-1} b_k$ has $b_k > 0$ , $b_{k+1} \le b_k$ for all $k \ge 1$ , and $b_k \to 0$ , then

\lvert S - s_n \rvert \le b_{n+1},

and $S - s_n$ has the sign of the first omitted term.

In plain words For a well behaved alternating series, the error is at most the size of the next term.

Example For $\sum \frac{(-1)^{k-1}}{k^2}$ with $n = 10$ , the bound is $b_{11} = \frac{1}{121} \approx 0.00826$ . In fact $s_{10} \approx 0.817962$ and $S = \frac{\pi^2}{12} \approx 0.822467$ , an error of about $0.00450$ . Since the first omitted term is positive, $s_{10}$ lies below $S$ .

Certified number of terms

Definition Given a tolerance $\varepsilon > 0$ , a number of terms $n$ is certified if a proven bound $B_n$ on the error, such as one of the three above, satisfies $B_n \le \varepsilon$ . Then the error is guaranteed to be at most $\varepsilon$ . The smallest certified $n$ depends on the bound used, and it can be larger than the number of terms that actually achieve the tolerance.

In plain words Watching the partial sums stop changing on the screen proves nothing. A certificate is a bound you have proved that is already below the tolerance.

Example Suppose you want $\sum \frac{1}{n^2}$ within $0.01$ . Using $s_n$ alone, the integral bound gives

R_n \le \int_n^{\infty} x^{-2}\,dx = \frac{1}{n},

so $n = 100$ terms are certified. Using the midpoint of the integral bounds instead, the error is at most

\frac{1}{2}\left(\frac{1}{n} - \frac{1}{n + 1}\right) = \frac{1}{2n(n + 1)},

which is at most $0.01$ once $n(n + 1) \ge 50$ , so $n = 7$ terms are certified. Indeed the estimate

s_7 + \frac{1}{2}\left(\frac{1}{8} + \frac{1}{7}\right) \approx 1.645726

is within $0.0008$ of $\frac{\pi^2}{6}$ .

Selecting a convergence test

Definition Selecting a convergence test means reading the form of $a_n$ and choosing a test whose hypotheses are easy to check for that form. No single test decides every series. The table lists the usual matches, roughly in the order to ask about them.

What you see in $a_n$	Test to try first
$a_n \not\to 0$	nth term test, the series diverges
$ar^n$	geometric series, converges exactly when $\lvert r \rvert < 1$
$\frac{1}{n^p}$	p-series, converges exactly when $p > 1$
$b_n - b_{n+1}$	telescoping
rational or algebraic in $n$	limit comparison with a p-series
factorials or $c^n$	ratio test
$(b_n)^n$	root test
$(-1)^n b_n$ with $b_n > 0$	absolute convergence first, then the alternating series test
$f(n)$ with $\int f$ easy	integral test

In plain words Look at the shape of the terms before computing anything. First rule out divergence by the nth term test, then match the form to the test that handles it best.

Example For $\sum \frac{n^3}{3^n}$ the exponential points to the ratio test, and

\frac{a_{n+1}}{a_n} = \frac{(n + 1)^3}{3^{n+1}} \cdot \frac{3^n}{n^3} = \frac{(n + 1)^3}{3n^3} \to \frac{1}{3} < 1,

so it converges. For $\sum \frac{1}{n^2 + \sqrt{n}}$ the terms are algebraic, and $\frac{1}{n^2 + \sqrt{n}} \le \frac{1}{n^2}$ settles it by direct comparison. Trying the ratio test on the second series would waste time, since its limit is $1$ .

Check yourself

How many terms of $\sum \frac{(-1)^{n-1}}{n^4}$ guarantee an error less than $10^{-4}$ ?

Show the answer

The sizes $\frac{1}{n^4}$ are positive, decreasing and tend to $0$ , so the alternating estimate applies. We need $\frac{1}{(n + 1)^4} < 10^{-4}$ , that is $n + 1 > 10$ . So $n = 10$ terms suffice. With $n = 9$ the bound is exactly $10^{-4}$ , which is not strictly less.

Bound the error of $\sum_{k=0}^{4} \frac{1}{2^k k!}$ as an estimate of $\sum_{k=0}^{\infty} \frac{1}{2^k k!} = \sqrt{e}$ .

Show the answer

For $k \ge 5$ the ratio of consecutive terms is $\frac{1}{2(k + 1)} \le \frac{1}{12}$ . The first omitted term is $\frac{1}{2^5 \cdot 5!} = \frac{1}{3840}$ , so the geometric tail bound gives

0 < \sqrt{e} - \sum_{k=0}^{4} \frac{1}{2^k k!} \le \frac{1/3840}{1 - 1/12} = \frac{1}{3520} \approx 0.00028.

Which test would you try first on $\sum \frac{2^n}{n!}$ and what does it give?

Show the answer

The factorial points to the ratio test. The ratio is $\frac{2}{n + 1} \to 0 < 1$ , so the series converges absolutely.

Estimates of Sums and Test Selection

Key terms

Absolute error

Integral bounds

Geometric tail bounds

Alternating error estimates

Certified number of terms

Selecting a convergence test

Check yourself

How this lesson connects

Builds on