The Integral Test for Positive Series of Real Numbers

One very useful test for convergence or divergence of a positive series of real numbers is called the integral test. It reduces the problem of series convergence to improper integral convergence when applicable. We state the result below.

Theorem 1: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be a positive series. If there exists a positive function $f$ that decreases and converges to $0$ on $[1, \infty)$ such that $f(n) = a_n$ for all $n \in \mathbb{N}$ then: a) $\displaystyle{\int_{1}^{n} f(x) \: dx + a_n \leq s_n \leq \int_{1}^{n} f(x) \: dx + a_1}$. b) $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges if and only if $\displaystyle{\int_1^{\infty} f(x) \: dx}$ converges. c) $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges if and only if $\displaystyle{\int_{1}^{\infty} f(x) \: dx}$ diverges.

• Proof of a) Fix $n \in \mathbb{N}$. Then the integral of $f$ on the interval $[1, n]$ is equal to the sum of the integrals of $f$ on the subintervals $[1, 2]$, $[2, 3]$, …, $[n - 1, n]$. In other words:

(1)

\begin{align} \quad \int_1^n f(x) \: dx = \sum_{k=1}^{n - 1} \left ( \int_k^{k+1} f(x) \: dx \right ) \end{align}

• Since $f$ is a decreasing function on $[1, \infty)$ we have that for all $k \in \{ 2, 3, ... \}$ and for all $x \in [x_{k-1}, x_k]$ that:

(2)

\begin{align} \quad f(k + 1) \leq f(x) \leq f(k) \quad (*) \end{align}

• Using $(*)$ we see that:

(3)

\begin{align} \quad \sum_{k=1}^{n - 1} \left ( \int_{k}^{k+1} f(k+1) \: dx \right ) & \leq \int_1^n f(x) \: dx \leq \sum_{k=1}^{n-1} \left ( \int_{k}^{k+1} f(k) \: dx \right ) \\ \quad \sum_{k=1}^{n - 1} \left ( \int_k^{k+1} a_{k+1} \: dx \right ) & \leq \int_1^n f(x) \: dx \leq \sum_{k=1}^{n-1} \left ( \int_k^{k+1} a_k \: dx \right ) \quad \end{align}

• Note that $\displaystyle{\int_k^{k+1} a_k \: dx} = a_k([k + 1] - k] = a_k$ for all $k$, and so the inequality above can be simplified to:

(4)

\begin{align} \quad \sum_{k=1}^{n - 1} a_{k+1} & \leq \int_1^n f(x) \: dx \leq \sum_{k=1}^{n - 1} a_k \\ \quad (a_2 + a_3 + ... + a_n) & \leq \int_1^n f(x) \: dx \leq (a_1 + a_2 + ... + a_{n-1} ) \\ \quad s_n - a_1 & \leq \int_1^n f(x) \: dx \leq s_n - a_n \\ \end{align}

• Since $\displaystyle{s_n - a_1 \leq \int_1^n f(x) \: dx}$ we see that $\displaystyle{s_n \leq \int_1^n f(x) \: dx + a_1}$. Furthermore, since $\displaystyle{\int_1^n f(x) \: dx \leq s_n - a_n}$ we see that $\displaystyle{\int_1^n f(x) \: dx + a_n \leq s_n}$. Combining these gives us that:

(5)

\begin{align} \quad \int_1^n f(x) \: dx + a_n \leq s_n \leq \int_1^n f(x) \: dx + a_1 \quad \blacksquare \end{align}

• Proof of b) $\Rightarrow$ Suppose that $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges, say to the sum $s$. Then the sequence of partial sums of the series, $(s_n)_{n=1}^{\infty}$ converges, say to $s$. Using part (a) above we have that for all $n \in \mathbb{N}$ that then:

(6)

\begin{align} \quad \int_1^n f(x) \: dx + a_n \leq s \end{align}

• Since $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges, we have that $\displaystyle{\lim_{n \to \infty} a_n = 0}$ and so:

(7)

\begin{align} \quad \int_1^{\infty} f(x) \: dx = \lim_{n \to \infty} \int_1^n f(x) \: dx \leq s \end{align}

• This shows that $\displaystyle{\int_1^{\infty} f(x) \: dx}$ converges since $f$ is a positive function.

• $\Rightarrow$ Suppose note that $\displaystyle{\int_1^{\infty} f(x) \: dx}$ converges, say to $s$. Then using part (a) again we have that:

(8)

\begin{align} \quad s_n \leq \int_1^{n} f(x) \: dx + a_1 \leq \int_1^{\infty} f(x) \: dx + a_1 = a_1 + s \end{align}

• Taking the limit as $n to \infty$ shows that $\displaystyle{\lim_{n \to \infty} s_n \leq a_1 + s}$. So the sequence of partial sums $(s_n)_{n=1}^{\infty}$ is bounded above. But $(a_n)_{n=1}^{\infty}$ is a positive sequence of numbers, so $(s_n)_{n=1}^{\infty}$ is strictly increasing. So, an increasing sequence of numbers that is bounded above is convergent, i.e., $(s_n)_{n=1}^{\infty}$ converges. So $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges. $\blacksquare$

• Proof of c) The statement is the contrapositive of (b). $\blacksquare$