The p-Series and Estimating Series Value (2024)

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]

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.

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].

The p-Series and Estimating Series Value (1)

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].

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

The p-Series and Estimating Series Value (2024)

FAQs

What is the p value of a series? ›

A p-value measures the probability of obtaining the observed results, assuming that the null hypothesis is true. The lower the p-value, the greater the statistical significance of the observed difference. A p-value of 0.05 or lower is generally considered statistically significant.

How do you find the value of P series? ›

What is the P series formula? All p-series can be represented as the infinite sum of the reciprocal powers of the natural numbers. The general formula is the sum of 1/n^p for n between 1 and infinity.

What is the P series test for series? ›

A test exists to describe the convergence of all p-series. That test is called the p-series test, which states simply that: If p > 1, then the series converges, If p ≤ 1, then the series diverges.

What is the equation for the P series? ›

1 np =1+ 1 2p + 1 3p + ... + 1 np + ... is called the p-series. Its sum is finite for p > 1 and is infinite for p ≤ 1.

How do I calculate the p-value? ›

  1. For a lower-tailed test, the p-value is equal to this probability; p-value = cdf(ts).
  2. For an upper-tailed test, the p-value is equal to one minus this probability; p-value = 1 - cdf(ts).

What is the rule for p series? ›

Theorem 7 (p-series). A p-series X 1 np converges if and only if p > 1. Proof. If p ≤ 1, the series diverges by comparing it with the harmonic series which we already know diverges.

How do you determine the value of P? ›

To find the p value for your sample, do the following:
  1. Identify the correct test statistic.
  2. Calculate the test statistic using the relevant properties of your sample.
  3. Specify the characteristics of the test statistic's sampling distribution.
  4. Place your test statistic in the sampling distribution to find the p value.

How do you calculate AP series? ›

The sum of 'n' terms in an AP can be calculated using a straightforward formula: Sn = n/2 [2a + (n-1)d]. Here, 'a' represents the first term, 'd' is the common difference, and 'n' denotes the number of terms.

How do you find the p-value of a sheet? ›

2. How to find a p-value using the T. TEST function
  1. Input your data samples into an Excel spreadsheet.
  2. Gather the number of tails and the type of t-test you want to perform.
  3. Use the formula =T. TEST(array 1, array 2, tails, type.)

What is P series also known as? ›

𝑝-series is a family of series where the terms are of the form 1/(nᵖ) for some value of 𝑝. The Harmonic series is the special case where 𝑝=1. These series are very interesting and useful.

What is the p-value in time series? ›

The p-value represents a probability of the error when expecting, that the trend differs from zero (i.e. probability, that there is no time change and the value is based on random fluctuations only).

What is the P formula? ›

The P-value formula provides for the equation that helps find out whether the hypothesis is correct or incorrect. It indicates the probability of a result to get exactly repeated or being similar in traits to the actual observation.

What is the series formula? ›

The 'nth' term of this arithmetic sequence, represented as 'an', can be computed using the formula: an = a + (n – 1) d. The total sum of the arithmetic series, denoted as 'Sn', can be calculated through the formula: Sn = n/2 (2a + (n – 1) d) (or) Sn = n/2 (a + an).

What is the value of p Fund series? ›

Pfund Series (nl=5)

Thus, the series is named after him. Pfund series is displayed when electron transition takes place from higher energy states(nh=6,7,8,9,10,…) to nl=5 energy state. All the wavelength of Pfund series falls in Infrared region of the electromagnetic spectrum.

What is the p-value for stationary series? ›

A p-value below a threshold (such as 5% or 1%) suggests we reject the null hypothesis (stationary), otherwise a p-value above the threshold suggests we fail to reject the null hypothesis (non-stationary). p-value > 0.05: Fail to reject the null hypothesis (H0), the data has a unit root and is non-stationary.

What is the p-value in sequencing? ›

p-value is a measure of how likely you are to get this data if no real difference existed. So, small p-value means your results are not by chance. But q-values are adjusted p-values by FDR approach, which show you even if your data is statistically significant, what would be False Discovery Rate.

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