# The Integral Test for Positive Series

We will now look at a technique for determining whether an **ultimately positive and ultimately non-increasing series** is convergence or divergent. We say that the series $\sum_{n=1}^{\infty} a_n$ is ultimately positive if for some $n \in \mathbb{N}$, then if $n ≥ N$ then $a_n ≥ 0$. The integral test for positive non-increasing series is as follows.

Theorem 1: Integral Test for Positive Series: Let $f(n) = a_n$ be a continuous ultimately non-increasing positive function for the interval $[N, \infty)$ where $N \in \mathbb{N}$. If $\int_{N}^{\infty} f(x) \: dx$ converges then $\sum_{n=N}^{\infty} a_n$ converges. Similarly, if $\int_{N}^{\infty} f(x) \: dx$ diverges to infinity then $\sum_{n=N}^{\infty} a_n$ diverges to infinity. |

We note that since our series is positive it cannot possibly diverge to negative infinity, and oscillating sequences can't be evaluated using this test.

**Proof of Theorem:**Let $\sum_{n=1}^{\infty} a_n$ be a series where the $n^{\mathrm{th}}$ partial sum $s_n = a_1 + a_2 + ... + a_n$. Now suppose that $n > N$. We can write our $n^{\mathrm{th}}$ partial sum as follows noting that $a_1 = f(1)$, $a_2 = f(2)$, …, $a_n = f(n)$:

- We note that $f(N+1), f(N+2), ..., f(n)$ represent the areas of the rectangles as illustrated below. For example, the first rectangles' area is calculated as $A = \mathrm{base} \times \mathrm{height} = 1 \cdot f(N+1)$. We thus note that $\sum_{n=N}^{\infty} a_n$ is an underestimation of the area bounded between $f$ and the $x$-axis over the interval $[N, \infty)$, and so:

- Now suppose that $\int_{N}^{\infty} f(x) \: dx$ was convergent, that is $\int_{N}^{\infty} f(x) \: dx = M$ for some $M \in \mathbb{R}$ such that $M > 0$. Therefore:

- Now we know that $s_n$ is bounded above. We also note that the sequence of partial sums is increasing since the sequences of terms are all positive. By the monotonic sequence theorem it follows that then $\sum_{n=N}^{\infty} a_n$ is convergent.

- Now we will look at showing that if the integral is divergent then the series must also be divergent.

- We note that the $n^{\mathrm{th}}$ partial sum $s_n = s_{N-1} + a_N + a_{N+1} + a_{N+2} + ... + a_n$ and so it follows that $s_n = s_{N-1} + f(N) + f(N+1) + f(N+2) + ... + f(n)$ which this time overestimates the area between the curve $f$ and the $x$-axis as illustrated:

So we get that $s_n ≥ s_{N-1} + \int_{N}^{\infty} f(x) \: dx$. If $\int_{N}^{\infty} f(x) \: dx$ diverges to infinity, that is $\int_{N}^{\infty} f(x) \: dx = \infty$, then it follows that $s_n ≥ \infty$ since $s_{N-1}$ is a finite number, and so $\sum_{n=N}^{\infty} a_n$ must be divergent as well. $\blacksquare$

We will now look at some examples of applying the integral test to determine whether a positive series is convergent or divergent. We must note though that if a series is convergent, then this test does not tell us what the series converges to necessarily.

## Example 1

**Using the integral test for positive series determine whether the series $\sum_{n=1}^{\infty} \frac{n}{1 + n^2}$ is convergent or divergent.**

We first note that this is a positive series. Since we are taking $n > 0$, the numerator is always positive as well, the denominator is always positive. We also note that this series is decreasing, so we can apply the integral test to determine whether this series is convergent or divergent.

(4)We will need to apply u-substitution to evaluate this integral. Let $u = 1 + x^2$ so that $\frac{1}{2}du = x$ and so then:

(5)Therefore the series $\sum_{n=1}^{\infty} \frac{n}{1 + n^2}$ is divergent.

## Example 2

**Using the integral test for positive series determine whether the series $\sum_{n=6}^{\infty} \frac{1}{n \ln n}$ is convergent or divergent.**

Since the numerator is always $1$ and positive, and the denominator is always positive, and the sequence of terms is decreasing we can apply the integral test here.

(6)Now let $u = ln x$. Therefore $du = \frac{1}{x} \: dx$ and thus making these substitutions we get that:

(7)Therefore the series $\sum_{n=6}^{\infty} \frac{1}{n \ln n}$ is divergent.