The Divergence and Integral Tests: Learn It 4

Estimating the Value of a Series

When we know a series [latex]\displaystyle\sum_{n=1}^{\infty} a_n[/latex] converges, we often want to estimate its sum. We can approximate this sum using any finite partial sum [latex]\displaystyle\sum_{n=1}^{N} a_n[/latex] where [latex]N[/latex] is a positive integer.

The key question is: How good is this approximation?

When we approximate an infinite series with its [latex]N[/latex]th partial sum, we define the remainder as:

[latex]{R}_{N}=\displaystyle\sum _{n=1}^{\infty }{a}_{n}-\displaystyle\sum _{n=1}^{N}{a}_{n}[/latex]

This remainder [latex]R_N[/latex] represents the “error” in our approximation—how much we’re missing by stopping at the [latex]N[/latex]th term instead of continuing infinitely.

For series that satisfy the conditions of the integral test, we can estimate how large this remainder is.

remainder estimate from the integral test

Suppose [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] is a convergent series with positive terms. Suppose there exists a function [latex]f[/latex] satisfying the following three conditions:

  1. [latex]f[/latex] is continuous,
  2. [latex]f[/latex] is decreasing, and
  3. [latex]f\left(n\right)={a}_{n}[/latex] for all integers [latex]n\ge 1[/latex].

Let [latex]S_N[/latex] be the [latex]N[/latex]th partial sum. Then:

[latex]{S}_{N}+{\displaystyle\int }_{N+1}^{\infty }f\left(x\right)dx<\displaystyle\sum _{n=1}^{\infty }{a}_{n}<{S}_{N}+{\displaystyle\int }_{N}^{\infty }f\left(x\right)dx[/latex].

 

Remainder Estimate: The remainder [latex]R_N = \displaystyle\sum_{n=1}^{\infty} a_n - S_N[/latex] satisfies:

[latex]{\displaystyle\int }_{N+1}^{\infty }f\left(x\right)dx<{R}_{N}<{\displaystyle\int }_{N}^{\infty }f\left(x\right)dx[/latex].

 

This gives us upper and lower bounds on how much error we make when approximating the series with [latex]S_N[/latex].

Visualizing the Remainder Estimate

Figure 4 illustrates how the remainder estimate works.

This shows two graphs side by side of the same decreasing concave up function y = f(x) that approaches the x-axis in quadrant 1. Rectangles are drawn with a base of 1 over the intervals N through N + 4. The heights of the rectangles in the first graph are determined by the value of the function at the right endpoints of the bases, and those in the second graph are determined by the value at the left endpoints of the bases. The areas of the rectangles are marked: a_(N + 1), a_(N + 2), through a_(N + 4).
Figure 4. Given a continuous, positive, decreasing function [latex]f[/latex] and a sequence of positive terms [latex]{a}_{n}[/latex] such that [latex]{a}_{n}=f\left(n\right)[/latex] for all positive integers [latex]n[/latex], (a) the areas [latex]{a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\cdots <{\displaystyle\int }_{N}^{\infty }f\left(x\right)dx[/latex], or (b) the areas [latex]{a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\cdots >{\displaystyle\int }_{N+1}^{\infty }f\left(x\right)dx[/latex]. Therefore, the integral is either an overestimate or an underestimate of the error.

We can represent the remainder [latex]R_N = a_{N+1} + a_{N+2} + a_{N+3} + \cdots[/latex] as the sum of areas of rectangles.

From the visual comparison, we see that the total area of these rectangles is:

  • Bounded above by [latex]\int_N^{\infty} f(x)dx[/latex]
  • Bounded below by [latex]\int_{N+1}^{\infty} f(x)dx[/latex]

This gives us the inequalities:

[latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\cdots >{\displaystyle\int }_{N+1}^{\infty }f\left(x\right)dx[/latex]

and

[latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\cdots <{\displaystyle\int }_{N}^{\infty }f\left(x\right)dx[/latex].

Therefore:

[latex]{\displaystyle\int }_{N+1}^{\infty }f\left(x\right)dx<{R}_{N}<{\displaystyle\int }_{N}^{\infty }f\left(x\right)dx[/latex].

 

Since the infinite series equals the partial sum plus the remainder:

[latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}={S}_{N}+{R}_{N}[/latex],

 

We can substitute our bounds for [latex]R_N[/latex] to get bounds for the entire series:

[latex]{S}_{N}+{\displaystyle\int }_{N+1}^{\infty }f\left(x\right)dx<\displaystyle\sum _{n=1}^{\infty }{a}_{n}<{S}_{N}+{\displaystyle\int }_{N}^{\infty }f\left(x\right)dx[/latex].
This result gives you a “sandwich” estimate for your series. Calculate [latex]S_N[/latex] (which you can compute exactly), then add the lower and upper integral bounds to get a range where the true sum must lie. The integrals tell you either an overestimate or underestimate of your approximation error.

Consider the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}[/latex].

  1. Calculate [latex]{S}_{10}=\displaystyle\sum _{n=1}^{10}\frac{1}{{n}^{3}}[/latex] and estimate the error.
  2. Determine the least value of [latex]N[/latex] necessary such that [latex]{S}_{N}[/latex] will estimate [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}[/latex] to within [latex]0.001[/latex].