### Learning Outcomes

- Estimate the value of a series by finding bounds on its remainder term

## The *p*-Series

The harmonic series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex] and the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex] are both examples of a type of series called a *p*-series.

### Definition

For any real number [latex]p[/latex], the series

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}[/latex]

is called a ** p-series**.

We know the *p*-series converges if [latex]p=2[/latex] and diverges if [latex]p=1[/latex]. What about other values of [latex]p\text{?}[/latex] In general, it is difficult, if not impossible, to compute the exact value of most [latex]p[/latex] -series. However, we can use the tests presented thus far to prove whether a [latex]p[/latex] -series converges or diverges.

If [latex]p<0[/latex], then [latex]\frac{1}{{n}^{p}}\to \infty [/latex], and if [latex]p=0[/latex], then [latex]\frac{1}{{n}^{p}}\to 1[/latex]. Therefore, by the divergence test,

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}\text{ diverges if }p\le 0[/latex].

If [latex]p>0[/latex], then [latex]f\left(x\right)=\frac{1}{{x}^{p}}[/latex] is a positive, continuous, decreasing function. Therefore, for [latex]p>0[/latex], we use the integral test, comparing

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}\text{ and }{\displaystyle\int }_{1}^{\infty }\frac{1}{{x}^{p}}dx[/latex].

We have already considered the case when [latex]p=1[/latex]. Here we consider the case when [latex]p>0,p\ne 1[/latex]. For this case,

[latex]{\displaystyle\int }_{1}^{\infty }\frac{1}{{x}^{p}}dx=\underset{b\to \infty }{\text{lim}}{\displaystyle\int }_{1}^{b}\frac{1}{{x}^{p}}dx=\underset{b\to \infty }{\text{lim}}\frac{1}{1-p}{x}^{1-p}{|}_{1}^{b}=\underset{b\to \infty }{\text{lim}}\frac{1}{1-p}\left[{b}^{1-p}-1\right][/latex].

Because

[latex]{b}^{1-p}\to 0\text{ if }p>1\text{ and }{b}^{1-p}\to \infty \text{ if }p<1[/latex],

we conclude that

[latex]{\displaystyle\int _{1}^{\infty}} \dfrac{1}{x^{p}}dx = \Bigg\{ \begin{array}{c} \frac{1}{p-1}\text{ if }p>1\\ \infty \text{ if }p<1\end{array}[/latex]

Therefore, [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}[/latex] converges if [latex]p>1[/latex] and diverges if [latex]0<p<1[/latex].

In summary,

[latex]{\displaystyle\sum _{n=1}^{\infty}} \dfrac{1}{n^{p}} \bigg\{ \begin{array}{l} \text{ converges if }p>1\\ \text{ diverges if }p\le 1\end{array}[/latex]

### Example:Testing for Convergence of *p*-series

For each of the following series, determine whether it converges or diverges.

- [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{4}}[/latex]
- [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{\frac{2}{3}}}[/latex]

Show Solution

### try it

Does the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{\frac{5}{4}}}[/latex] converge or diverge?

Hint

Show Solution

Watch the following video to see the worked solution to the above Try It.

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You can view the transcript for this segmented clip of “5.3.4” here (opens in new window).

## Estimating the Value of a Series

Suppose we know that a series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum [latex]\displaystyle\sum _{n=1}^{N}{a}_{n}[/latex] where [latex]N[/latex] is any positive integer. The question we address here is, for a convergent series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex], how good is the approximation [latex]\displaystyle\sum _{n=1}^{N}{a}_{n}\text{?}[/latex] More specifically, if we let

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

be the remainder when the sum of an infinite series is approximated by the [latex]N\text{th}[/latex] partial sum, how large is [latex]{R}_{N}\text{?}[/latex] For some types of series, we are able to use the ideas from the integral test to estimate [latex]{R}_{N}[/latex].

### theorem: 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:

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

Let [latex]{S}_{N}[/latex] be the *N*th partial sum of [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex]. For all positive integers [latex]N[/latex],

[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].

In other words, the remainder [latex]{R}_{N}=\displaystyle\sum _{n=1}^{\infty }{a}_{n}-{S}_{N}=\displaystyle\sum _{n=N+1}^{\infty }{a}_{n}[/latex] satisfies the following estimate:

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

This is known as the remainder estimate.

We illustrate the Remainder Estimate from the Integral Test in Figure 4. In particular, by representing the remainder [latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\cdots [/latex] as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by [latex]{\displaystyle\int }_{N}^{\infty }f\left(x\right)dx[/latex] and bounded below by [latex]{\displaystyle\int }_{N+1}^{\infty }f\left(x\right)dx[/latex]. In other words,

[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].

We conclude that

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

Since

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

where [latex]{S}_{N}[/latex] is the [latex]N\text{th}[/latex] partial sum, we conclude that

[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].

### Example:Estimating the Value of a Series

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

- Calculate [latex]{S}_{10}=\displaystyle\sum _{n=1}^{10}\frac{1}{{n}^{3}}[/latex] and estimate the error.
- 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].

Show Solution

[latex]{S}_{10}=1+\frac{1}{{2}^{3}}+\frac{1}{{3}^{3}}+\frac{1}{{4}^{3}}+\cdots +\frac{1}{{10}^{3}}\approx 1.19753[/latex].

By the remainder estimate, we know

[latex]{R}_{N}<{\displaystyle\int }_{N}^{\infty }\frac{1}{{x}^{3}}dx[/latex].

We have

[latex]{\displaystyle\int _{10}^{\infty }}\frac{1}{{x}^{3}}dx= {\underset{b\to \infty }\lim} {\displaystyle\int _{10}^{b}} \frac{1}{{x}^{3}}dx= {\underset{b\to\infty}\lim} \left[ -\frac{1}{2x^2} \right] _{N}^{b}= {\underset{b\to \infty }\lim} \left[-\frac{1}{2{b}^{2}} + \frac{1}{2{N}^{2}}\right]= \frac{1}{2{N}^{2}}[/latex].

Therefore, the error is [latex]{R}_{10}<\frac{1}{2{\left(10\right)}^{2}}=0.005[/latex].

- Find [latex]N[/latex] such that [latex]{R}_{N}<0.001[/latex]. In part a. we showed that [latex]{R}_{N}<\frac{1}{2{N}^{2}}[/latex]. Therefore, the remainder [latex]{R}_{N}<0.001[/latex] as long as [latex]\frac{1}{2{N}^{2}}<0.001[/latex]. That is, we need [latex]2{N}^{2}>1000[/latex]. Solving this inequality for [latex]N[/latex], we see that we need [latex]N>22.36[/latex]. To ensure that the remainder is within the desired amount, we need to round up to the nearest integer. Therefore, the minimum necessary value is [latex]N=23[/latex].

### try it

For [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{4}}[/latex], calculate [latex]{S}_{5}[/latex] and estimate the error [latex]{R}_{5}[/latex].

Hint

Show Solution