The Integral Test For Series

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One way we can determine whether or not a series converges is to view it as an area and see if the area is finite. Of course, we know how to find the area under a function (by integrating), so this suggests that integration might somehow be useful. That’s what’s involved in the integral test for series.

Suppose we want to know whether the series

    \[ \sum_{n=1}^{\infty} a_n \]

converges or not. Define a function f(x) by replacing all the n’s in a_n with x’s. If f is continuous, positive, and decreasing, and if

    \[ \int_1^{\infty} f(x)\,dx \]

converges, so does the sum. Conversely, if the integral diverges, the sum does, too.

The integral test is very powerful when you can integrate the terms of the series you’re looking at. Let’s have an example: does the series

    \[ \sum_{n=1}^{\infty}\frac{1}{n^2} \]

converge?

Answer: we can apply the integral test. Define f(x)=\frac{1}{x^2}. Then a_n = f(n), f is continuous, positive, and decreasing on (1, \infty), and so the integral test applies. We have

    \begin{align*} \int_1^{\infty}\frac{1}{x^2}\,dx & = \lim_{t\rightarrow\infty}\int_1^t x^{-2}\,dx \\ & = \lim_{t\rightarrow\infty} \left.-x^{-1}\right|_1^t \\ & = \lim_{t\rightarrow\infty} -\frac{1}{t} + \frac{1}{1}\\ & = 1. \end{align*}

The integral converges, and therefore so does the series.

Another common example: does

    \[ \sum_{k=2}^{\infty}\frac{1}{k\ln k} \]

converge?

Answer: define f(x) = \frac{1}{x\ln x}. Then f is continuous and positive on (2,\infty). Notice that x\ln x is increasing on this interval, and so its reciprocal is decreasing. Therefore the integral test applies:

    \[ \int_2^{\infty} \frac{1}{x\ln x}\,dx & = \lim_{t\rightarrow\infty}\int_2^t \frac{1}{x\ln x}\, dx, \]

and now we can let u=\ln x in the integral, so that du = 1/x\,dx. With the new limits, the integral is

    \begin{align*} \lim_{t\rightarrow\infty} \int_{\ln 2}^{\ln t}\frac{1}{u}\,du & = \lim_{t\rightarrow\infty}\left.\ln |u|\right|_{\ln 2}^{\ln t} \\ & = \lim_{t\rightarrow\infty}\ln(\ln t) - \ln |\ln 2|\\ & = \infty. \end{align*}

Since the integral diverges, so does the given series.