Integral Test For Convergence

The integral test for convergence, a rather elegant piece of machinery in the realm of mathematics , is a method employed to ascertain the convergence of infinite series whose terms exhibit a monotonic quality. It’s a technique polished and presented by the likes of Colin Maclaurin and Augustin-Louis Cauchy , and as such, it occasionally bears their combined moniker, the Maclaurin–Cauchy test.

Statement of the Test

Imagine, if you will, an integer designated as N, and a function, let’s call it f, meticulously defined across the vast, unbounded interval stretching from N to infinity, denoted as [N, ∞). This function f is stipulated to be monotone decreasing throughout this domain. Under these conditions, the infinite series:

$$ \sum_{n=N}^{\infty} f(n) $$

will converge to a finite real number if and only if its integral counterpart, the improper integral :

$$ \int_{N}^{\infty} f(x) , dx $$

also yields a finite value. Conversely, should this integral diverge , meaning it doesn’t settle on a finite number, then the corresponding infinite series is doomed to diverge as well. It’s a rather definitive relationship, wouldn’t you agree?

Remark

The elegance of this test isn’t merely in its dichotomy of convergence or divergence. If the improper integral happens to be finite, the proof itself furnishes us with valuable lower and upper bounds for the sum of the infinite series:

$$ \int_{N}^{\infty} f(x) , dx \leq \sum_{n=N}^{\infty} f(n) \leq f(N) + \int_{N}^{\infty} f(x) , dx $$

This is quite handy, really. It gives you a tangible range for the series’ sum, should it converge.

Now, a subtle point: if the function f(x) were found to be increasing rather than decreasing, one could simply consider its negative, -f(x). This transformed function would then be monotone decreasing, allowing the theorem to be applied without a hitch.

Some textbooks might insist on f being strictly positive. While that’s often the case in introductory examples, it’s not a strictly necessary condition for the test’s validity. If f is negative and decreasing, both the series $\sum f(n)$ and the integral $\int f(x) dx$ will diverge in tandem. It’s a bit like observing two ships sailing into the same storm; if one founders, the other is likely to follow.

Proof

The foundation of this proof rests on the comparison test . We meticulously compare each term $f(n)$ of the series with the integral of the function f over specific intervals: either $[n-1, n)$ or $[n, n+1)$.

First, we must establish that our monotone function $f$ is continuous “almost everywhere.” This might sound like a lawyer’s loophole, but it’s a crucial technicality. For every point $x$ where $f$ might falter and exhibit discontinuity, we can find a rational number $c(x)$ nestled within the range of values that $f$ approaches around $x$. This process, mapping each point of discontinuity to a unique rational number, demonstrates that the set of discontinuities, $D$, is countable . A function that is continuous almost everywhere is, for our purposes, Riemann integrable .

Given that $f$ is a monotone decreasing function, a couple of inequalities naturally arise for any integer $n \ge N$:

$$ f(x) \leq f(n) \quad \text{for all} \ x \in [n, \infty) $$

and

$$ f(n) \leq f(x) \quad \text{for all} \ x \in [N, n] $$

From these, we can derive the following crucial inequalities:

For every integer $n \ge N$:

$$ \int_{n}^{n+1} f(x) , dx \leq \int_{n}^{n+1} f(n) , dx = f(n) $$

And for every integer $n \ge N+1$:

$$ f(n) = \int_{n-1}^{n} f(n) , dx \leq \int_{n-1}^{n} f(x) , dx $$

Now, if we sum these inequalities from $n=N$ up to some larger integer $M$, we start to see the structure emerge. From the first inequality, summing over $n$ from $N$ to $M$:

$$ \int_{N}^{M+1} f(x) , dx = \sum_{n=N}^{M} \int_{n}^{n+1} f(x) , dx \leq \sum_{n=N}^{M} f(n) $$

And from the second inequality, with a slight adjustment for the initial term $f(N)$:

$$ \sum_{n=N}^{M} f(n) = f(N) + \sum_{n=N+1}^{M} f(n) \leq f(N) + \sum_{n=N+1}^{M} \int_{n-1}^{n} f(x) , dx = f(N) + \int_{N}^{M} f(x) , dx $$

Combining these two results, we arrive at the bounds mentioned earlier:

$$ \int_{N}^{M+1} f(x) , dx \leq \sum_{n=N}^{M} f(n) \leq f(N) + \int_{N}^{M} f(x) , dx $$

As $M$ relentlessly marches towards infinity, these inequalities solidify into the core statement of the integral test. The convergence or divergence of the integral directly dictates the fate of the series.

Applications

Consider the venerable harmonic series :

$$ \sum_{n=1}^{\infty} \frac{1}{n} $$

This series, famously, diverges. We can see this by applying the integral test. The integral of $1/x$ from $1$ to $M$ is $\ln M$. As $M$ approaches infinity, $\ln M$ also approaches infinity.

$$ \int_{1}^{M} \frac{1}{n} , dn = \ln n \Big|_{1}^{M} = \ln M \to \infty \quad \text{for} \ M \to \infty $$

Since the integral diverges, so too must the harmonic series.

Now, let’s look at a related series, one involving the Riemann zeta function :

$$ \zeta(1+\varepsilon) = \sum_{n=1}^{\infty} \frac{1}{n^{1+\varepsilon}} $$

This series converges for any $\varepsilon > 0$. Again, the integral test confirms this. The integral of $1/n^{1+\varepsilon}$ from $1$ to $M$ is finite:

$$ \int_{1}^{M} \frac{1}{n^{1+\varepsilon}} , dn = \left. -\frac{1}{\varepsilon n^{\varepsilon}} \right|_{1}^{M} = \frac{1}{\varepsilon} \left( 1 - \frac{1}{M^{\varepsilon}} \right) \leq \frac{1}{\varepsilon} < \infty \quad \text{for all} \ M \geq 1 $$

The integral is bounded, so the series converges. Furthermore, the bounds derived from the test give us an upper estimate for the zeta function:

$$ \zeta(1+\varepsilon) = \sum_{n=1}^{\infty} \frac{1}{n^{1+\varepsilon}} \leq \frac{1+\varepsilon}{\varepsilon} $$

This provides a useful comparison point for understanding the behavior of the Riemann zeta function near its pole at $s=1$.

The Borderline Between Divergence and Convergence

The examples of the harmonic series and the $\zeta(1+\varepsilon)$ series naturally lead to a fascinating question: are there sequences $f(n)$ that decrease to zero “slower” than $1/n^{1+\varepsilon}$ but “faster” than $1/n$? Specifically, sequences where:

$$ \lim_{n\to \infty} \frac{f(n)}{1/n} = 0 \quad \text{and} \quad \lim_{n\to \infty} \frac{f(n)}{1/n^{1+\varepsilon}} = \infty $$

for any $\varepsilon > 0$? And do the corresponding series $\sum f(n)$ still diverge? This line of inquiry allows us to probe the very edge, the “borderline,” between divergence and convergence.

Using the integral test, mathematicians have demonstrated that for any positive natural number $k$, the series:

$$ \sum_{n=N_{k}}^{\infty} \frac{1}{n\ln(n)\ln _{2}(n)\cdots \ln _{k-1}(n)\ln _{k}(n)} $$

still diverges. (This is a generalization of the proof that the sum of the reciprocals of the primes diverges, which corresponds to $k=1$). However, the slightly modified series:

$$ \sum_{n=N_{k}}^{\infty} \frac{1}{n\ln(n)\ln _{2}(n)\cdots \ln _{k-1}(n)(\ln _{k}(n))^{1+\varepsilon}} $$

converges for every $\varepsilon > 0$. Here, $\ln_k(x)$ denotes the $k$-fold composition of the natural logarithm, defined recursively.

To establish the divergence of the first series, we observe that the derivative of $\ln_{k+1}(x)$ is $\frac{1}{x\ln(x)\cdots \ln_{k}(x)}$. Therefore, the integral of this form from $N_k$ to infinity diverges to infinity.

For the convergence of the second series, we utilize the power rule and the chain rule. The derivative of $-\frac{1}{\varepsilon (\ln _{k}(x))^{\varepsilon}}$ is precisely the integrand of the second series. This means the integral of the second series’ integrand from $N_k$ to infinity is finite, thus confirming its convergence by the integral test. These constructions reveal an astonishingly fine-grained hierarchy of divergence and convergence, a testament to the subtle nuances of infinite sums.