# The Formula

If f is continuous and decreasing, and $$f(x)\geq0$$
for x $$\geq 1$$, then $$\int\limits_{1}^{\infty}f(x)dx$$ and $$\sum\limits_{n=1}^{\infty}a_{n}$$ either both converge or both diverge.

# Notes

• In order to use the Integral Test, when you transform the series into an integral, you have to make sure that between the limits of integration the function is continuous, positive and decreasing
• The value you get after integrating is not what the series converges to. The only significance of the number you get is to tell you whether answer was finite or not. The number itself has no meaning. Do not make the mistake of thinking that the series converges to that number.
• The integral test is useful when you are presented with a series that contains a function that you are able to integrate
• It takes practice to get comfortable recognizing which series are right for this test

# Examples

$\text{ Example 1)} \sum\limits_{n=2}^{\infty}\frac{ln(n)}{n}$
 $$\int\limits_{2}^{\infty}\frac{ln(x)}{x}$$ $$\text{from } 2 \text{ to } \infty$$ the function is continuous, decreasing and positive

$$\int\limits_{2}^{\infty}\frac{ln(x)}{x}=\lim\limits_{a\rightarrow\infty}\int\limits_{2}^{a}\frac{ln(x)}{x} =\lim\limits_{a\rightarrow\infty}\int\limits_{ln(2)}^{a}u=\frac{u^2}{2}\big|_{ln(2)}^\infty=\infty$$
Therefore, this series diverges.
$\text{ Example 2)} \sum\limits_{n=1}^{\infty}\frac{tan^{-1}n}{1+n^2}$
 $$\int\limits_{1}^{\infty}\frac{tan^{-1}x}{1+x^2}$$ from 1 to $$\infty$$ the function is continuous, decreasing and positive

$$\int\limits_{1}^{\infty}\frac{tan^{-1}x}{1+x^2}=\lim\limits_{a\rightarrow\infty}\int\limits_{1}^{a}\frac{tan^{-1}x}{1+x^2}=\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}u \\ =\frac{u^2}{2}\big|_{\frac{\pi}{4}}^\frac{\pi}{2}=\frac{(\frac{\pi}{2})^2}{2}-\frac{(\frac{\pi}{4})^2}{2}$$
Therefore, this series converges.