P-Series Convergence and Divergence


Lesson overview

In this lesson, we'll prove when the value of the p-series \(\sum_{n=1}^∞\frac{1}{n^p}\) converges to a finite value and when its diverges to infinity. We'll show that when \(0<p<1\), the p-series diverges; and when \(p>1\), the p-series converges.

                      Figure 1 (click to enlarge)

A graph of the function \(y=\frac{1}{x^p}\) is shown in Figure 1 and in the video.\(^{[1]}\) The first term in the p-series is \(\frac{1}{1^p}\)  and is the area of the first rectangle. The second term in this p-series is \(\frac{1}{2^p}\)  and is the area of the second rectangle. The sum of all the terms in the p-series is equal to the sum of all the areas of each rectangle and is the upper-Riemann estimate of the area under the curve \(y=1/x^p\)  from \(x=1\) to \(x=∞\). The integral \(∫_1^∞\frac{1}{x^p}dx\) is equal to the actual area under the curve from \(x=1\) to \(x=∞\). For \(p>0\), the function \(y=\frac{1}{x^p}\) is always positive; therefore, the area underneath it, \(∫_1^∞\frac{1}{x^p}dx\), is always positive and the overestimate of that area, \(\sum_{n=1}^∞\frac{1}{n^p}\), is also always positive. Since the p-series is an overestimate of the actual area, it follows that \(∫_1^∞\frac{1}{x^p}dx<\sum_{n=1}^∞\frac{1}{n^p}\). The expression \(1+∫_1^∞\frac{1}{x^p}dx\) equals the area of the shaded regions (region 1 and region 2) in Figure 2. Since the p-series equals the area of region 1 plus the lower-Riemann estimate of region 2, it follows that \(\sum_{n=1}^∞\frac{1}{n^p}<1+∫_1^∞\frac{1}{x^p}dx\) and, thus,

                 Figure 2 (click to enlarge)

From Inequalities (1), we see that if \(∑_{n=1}^∞\frac{1}{n^p}\) converges (that is, equals a finite value), then since \(∫_1^∞\frac{1}{x^p}dx\) must be positive but less than \(\sum_{n=1}^∞\frac{1}{n^p}\) (which is a finite value if it converges) the integral \(∫_1^∞\frac{1}{x^p}dx\) must also be equal to some finite value and converge. We can also make the converse statement: if \(∫_1^∞\frac{1}{x^p}dx\) converges then \(1+∫_1^∞\frac{1}{x^p}dx\) must also equal a finite value and converge and thus the value of the p-series is “in between” two finite values (given by \(∫_1^∞\frac{1}{x^p}dx\) and \(1+∫_1^∞\frac{1}{x^p}dx\)) and must also converge. We can summarize both of these statement by saying,

$$\text{For }p>0,\text{ the p-series }\sum_{n=1}^∞\frac{1}{n^p}\text{ converges if and only if the integral }∫_1^∞\frac{1}{x^p}dx\text{ converges.}$$

If the integral \(∫_1^∞\frac{1}{x^p}dx\) diverges (that is, equals infinity), then since \(\sum_{n=1}^∞\frac{1}{n^p}>∫_1^∞\frac{1}{x^p}dx\) it follows that the p-series \(∑_{n=1}^∞\frac{1}{n^p}\)  is also infinite and diverges. We can once again make the converse statement: if the p-series \(∑_{n=1}^∞\frac{1}{n^p}\) diverges, then since \(1+∫_1^∞\frac{1}{x^p}dx>∑_{n=1}^∞\frac{1}{n^p}\), it follows that \(1+∫_1^∞\frac{1}{x^p}dx\) must also be infinite. If we subtract one from an infinite value to get \(∫_1^∞\frac{1}{x^p}dx\), we’ll still end up with an infinite value. Thus,
$$\text{For }p>0,\text{ the p-series }\sum_{n=1}^∞\frac{1}{n^p}\text{ diverges if and only if the integral }∫_1^∞\frac{1}{x^p}dx\text{ diverges.}$$

In other words if the p-series converges/diverge we know that the integral converges/diverges, and vice versa. Let’s now see for what values of \(p\) (greater than zero) there is convergence and for what values of \(p\) there is divergence. We’ll prove that for values of \(p\) within the range \(0<p≤1\), both the integral and p-series converges and that for values of \(p\) within the range \(p>1\) both the integral and p-series diverges.

First, let’s see what happen if \(p=1\). We can rewrite the improper integral \(∫_1^∞\frac{1}{x^p}dx\) as \(∫_1^∞\frac{1}{x^p}dx=\lim_{m→∞}⁡∫_1^m\frac{1}{x^p}dx\). If \(p=1\). then \(∫_1^m\frac{1}{x^p}dx=∫_1^m\frac{1}{x}dx=[ln⁡|x| ]_1^m=ln⁡|m|-ln⁡(1)\). Since we must raise \(e\) to the power zero to get one, \(ln⁡(1)=0\). Thus, \(∫_1^m\frac{1}{x}dx=ln⁡|m|\) and \(∫_1^∞\frac{1}{x^p}dx=\lim_{m→∞}⁡∫_1^m\frac{1}{x^p}dx=lim_{m→∞}⁡ln⁡|m|\). The natural logarithm \(ln⁡|m|\) is the power \(e\) must be raised to in order to get \(|m|\); in other words, \(e^{ln⁡|m|}=|m|\). Clearly, if \(m→∞\), the power of \(e\) (which is \(ln⁡|m|\)) must also approach infinity. Therefore, if \(p=1\), then \(∫_1^∞\frac{1}{x^p}dx=lim_{m→∞}⁡ln⁡|m| =∞\) and the integral diverges; since the integral diverges it also follows that the p-series \(\sum_{n=1}^∞\frac{1}{n^p}\) diverges. In summary,
$$\text{If }p=1,\text{ the integral }∫_1^∞\frac{1}{x^p}dx\text{ and the p-series }\sum_{n=1}^∞\frac{1}{n^p}\text{ diverges.}$$

Let’s now see what happens to the integral \(∫_1^∞\frac{1}{x^p}dx\) and the p-series \(∑_{n=1}^∞\frac{1}{n^p}\) when \(0<p<1\) and when \(p>1\) but finite. Again we can rewrite the improper integral as \(∫_1^∞\frac{1}{x^p}dx=lim_{m→∞}⁡∫_1^m\frac{1}{x^p}dx\). To solve the anti-derivative \(∫_1^mx^{-p}dx\) we raise the exponent by one and then divide by the new exponent; in other words, \(∫_1^mx^{-p}dx=[\frac{x^{1-p}}{1-p}]_1^m=\frac{m^{1-p}}{1-p}-\frac{1^{1-p}}{1-p}\). Since one raised to any power always equals one, \(1^{1-p}=1\) and \(∫_1^mx^{-p}dx=\frac{m^{1-p}}{1-p}-\frac{1}{1-p}\). If we take the limit as \(m→∞\) on both sides, we get \(∫_1^∞\frac{1}{x^p}dx=lim_{m→∞}⁡∫_1^mx^{-p}dx=\frac{1}{1-p}lim_{m→∞}⁡(m^{1-p}-1)\). If \(p>1\), then the power \(1-p\) of \(m\) is negative; you can also view this as \(\frac{1}{m^{\text{positive number}}}\). In this case, as \(m→∞\), the quantity \(m^{1-p}=\frac{1}{m^{-(1-p)}}\) will approach zero and the integral converges. As discussed earlier if the integral converges then the p-series must also converge. Thus,

$$\text{If }p>1,\text{ then the integral }∫_1^∞\frac{1}{x^p}dx\text{ and the p-series }∑_{n=1}^∞\frac{1}{n^p}\text{ converges.}$$

If \(0<p<1\), then the exponent \(1-p\) of \(m\) is positive. If \(m→∞\), then the quantity \(m^{1-p}\) (with its positive exponent) will also go to infinity. Thus, if \(0<p<1\), then the integral \(∫_1^∞\frac{1}{x^p}dx\) diverges; since the integral diverges it necessarily follows that the p-series \(∑_{n=1}^∞\frac{1}{n^p}\)  also diverges. Thus,
    $$\text{If }0<p<1,\text{ then the integral }∫_1^∞\frac{1}{x^p}dx\text{ and the p-series }∑_{n=1}^∞\frac{1}{n^p}\text{ diverges.}$$

This article is licensed under a CC BY-NC-SA 4.0 license.


1. Khan Academy. "Proving p series convergence criteria". Online video clip. YouTube. YouTube, 16 December 2016. Web. 27 March 2017.