# 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.

This video was produced by the Khan Academy.$$^{[1]}$$

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,
$$∫_1^∞\frac{1}{x^p}dx<\sum_{n=1}^∞\frac{1}{n^p}<1+∫_1^∞\frac{1}{x^p}dx.\tag{1}$$

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,
$$\text{If }0<p<1,\text{ then the integral }∫_1^∞\frac{1}{x^p}dx\text{ and the p-series }∑_{n=1}^∞\frac{1}{n^p}\text{ diverges.}$$

References

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