23 - Integral Test

Integral Test

Let $f:[1,\infty )\to \mathbb{R}$ be a function
Assume that

1. $f$ is non-negative
   $f(x)\ge 0$ for all $x\in [1,\infty )$
2. $f$ is decreasing
   if $x<y$ then $f(x)\ge f(y)$
3. ${\int }_k^{k+1}f(x)dx$ exists for each $k\ge 1$

Then

Let $s_n={\sum }_{k=1}^{n}f(k)$
Let $I_n={\int }_1^{n}f(x)dx$

1. Let ${\sigma }_n=s_n-I_n$ then $({\sigma }_n)$ converges.
   Let ${\sigma }_n\to \sigma$ as $n\to \infty$ then

$$0\le \sigma \le f(1)$$

2. $\sum f(k)$ converges if and only if $(I_n)$ converges

3. If $f$ is continuous then ${\int }_k^{k+1}f(x)dx$ exists for each $k\ge 1$

Proof

1. Since $f$ is decreasing, for $x\in [k,k+1]$ then

$$f(k+1)\le f(x)\le f(k)$$

Hence

$$f(k+1)={\int }_k^{k+1}f(k+1)d\le {\int }_k^{k+1}f(x)dx\le {\int }_k^{k+1}f(k)dx=f(k)$$

So

$$\begin{array}{rl}f(2)& \le {\int }_2^{3}f(x)dx\le f(1)\\ f(3)& \le {\int }_2^{3}f(x)dx\le f(2)\\ & \text{and}\\ f(n)& \le {\int }_n^{n-1}f(x)dx\le f(n-1)\end{array}$$

Adding them we get

$$s_n-f(1)\le I_n\le s_n-f(n)$$

Therefore

$$0\le f(n)\le s_n-I_{}n\le f(1)$$

Therefore

$$0\le {\sigma }_n\le f(1)\text{ for all }n\ge 1$$

With

$$\begin{array}{rl}{\sigma }_{n+1}-{\sigma }_n& =s_{n+1}-I_{n+1}-s_n+I_n\\ & =f(n+1)-{\int }_n^{n+1}f(x)dx\le 0\end{array}$$

So $({\sigma }_n)$ is decreasing and bounded above
Hence by Monotone Sequence Theorem, it converges
Suppose ${\sigma }_n\to \sigma$ as $n\to \infty$
As limits preserve weak inequalities, and $0\le {\sigma }_n\le f(1)$ for all $n\ge 1$

$$0\le \sigma \le f(1)$$

2. If $(s_n)$ converges then by AOL so does $(I_n)$ as $I_n=s_n-{\sigma }_n$
   If $(I_n)$ converges then by AOL so does $(s_n)$ as $s_n=I_n+{\sigma }_n$