The p-Series and Estimating Series Value

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.

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

try it

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

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

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

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:

  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 Nth 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].

 

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.

Example: Estimating the Value of a Series

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].

[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].