Integral Test for Convergence

1. So we have the definition for a series to converge, but what about some slick tests?

Wait a moment: wasn’t a Riemann Sum an infinite series? Why don’t we use integration to test if a series converges or diverges?

We can do something like this. Lets investigate it further.

2. An infinite series could be thought of as \sum f(n) where f is a function of a real number…we just restrict attention to positive integers.

We can obviously test if f(x)\to 0 as x\to\infty using L’Hopital’s Rule. So we will assume that f(n+1)\leq f(n) for each n=1,2,\dots.

So what? Well, we see that
(1a)\displaystyle\int^{n+1}_{n} f(x)\,\mathrm{d}x \leq\int^{n+1}_{n} f(n)\,\mathrm{d}x =f(n)
and
(1b)\displaystyle f(n)=\int^{n}_{n-1} f(n)\,\mathrm{d}x \leq\int^{n}_{n-1} f(x)\,\mathrm{d}x.
It thus stands to reason that we can sum over n to get
(2)\displaystyle\int^{\infty}_{1} f(x)\,\mathrm{d}x \leq\sum^{\infty}_{n=1}f(n)\leq f(1)+\int^{\infty}_{1} f(x)\,\mathrm{d}x.
Consequently, the series converges if and only if the integral \int^{\infty}_{1}f(x)\,\mathrm{d}x is finite.

If the integral is infinite, the series diverges. This is quite clean, clever, and cute.

Example 1 (Harmonic Series). The classic example is the Harmonic series
(3)\displaystyle\sum^{\infty}_{n=1}\frac{1}{n}.
We see that the function f(x)=1/x has its integral be
(4)\displaystyle\int^{c}_{1}\frac{\mathrm{d}x}{x}=\ln(c)-\ln(1)=\ln(c).
Taking the limit as c\to\infty gives us f(c)\to\infty. By the integral test, the Harmonic Series diverges!

3. Usefulness. Consider the series
(5)\displaystyle\sum^{\infty}_{n=1}\frac{1}{n^{2}+4n}.
What does it converge to? Does it converge? Well, we see that the integral test gives us
(6)\displaystyle\begin{aligned}\int^{\infty}_{1}\frac{\mathrm{d}x}{x^{2}+4x}  &=\lim_{c\to\infty}\int^{c}_{1}\frac{\mathrm{d}x}{x^{2}+4x}\\  &=\lim_{c\to\infty}\int^{c}_{1}\left[\frac{1}{4x}-\frac{1}{4}\frac{1}{x+4}\right]\,\mathrm{d}x\\  &=\lim_{c\to\infty}\frac{1}{4}[\ln(x)-\ln(x+4)]^{c}_{1}\\  &=\lim_{c\to\infty}\frac{1}{4}\left[\ln\left(\frac{c}{c+4}\right)-\ln(1/5)\right]\\  &=\frac{-1}{4}\ln(1/5)=\frac{\ln(5)}{4}\end{aligned}
Thus we see the series converges.

Moreover, we can use the integral test to tell us how good using, e.g., the first eight terms would estimate the sum. We see that
(7)\displaystyle\begin{aligned}  \int^{\infty}_{9}\frac{\mathrm{d}x}{x(x+4)}  &\leq\mbox{the tail sum}  \leq\int^{\infty}_{8}\frac{\mathrm{d}x}{x(x+4)}\\  \frac{\ln(13/9)}{4}&\leq\mbox{the tail sum}  \leq\frac{\ln(3/2)}{4}\\  0.0919312&\leq\mbox{the tail sum}\leq0.1013663  \end{aligned}
So using only the first eight terms or so gives us an approximation good to a digit or so. But that’s something!

Theorem 1. The series \sum n^{-k}, where k is a fixed constant, converges if and only if k<1.

Proof. We use the integral test on f(x)=x^{-k}. We see if k=1, the integral test tells us the series diverges:
(8)\displaystyle \lim_{c\to\infty}\int^{c}_{1}\frac{\mathrm{d}x}{x}=\lim_{c\to\infty}\ln(c)=\infty
On the other hand, we see if k>1, we have the integral test tell us
(9)\displaystyle \lim_{c\to\infty}\int^{c}_{1}\frac{\mathrm{d}x}{x^{k}}=\lim_{c\to\infty}\frac{-c^{1-k}}{k+1}-\frac{-1}{k+1}=\frac{1}{k+1}
which implies the series converges.

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About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Analysis, Approximation, Calculus, Infinite Series. Bookmark the permalink.

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