1. ## Integral Test

infinity over Sigma n=1 1/n(n+1) Find whether it converges or diverges using the integral test. I know your suppose to use partial fractions but Im stuck on how to do that with an integral test.

infinity over Sigma n=1 1/n(n+1) Find whether it converges or diverges using the integral test. I know your suppose to use partial fractions but Im stuck on how to do that with an integral test.
Fir check that the test applies (if $a_n = f(n)$ is positive, continuous and decreasing)

then consider

$\int _1^{\infty} f(x)\,dx$

$\int _1^{\infty} \frac{1}{x(x+1)}\,dx = \lim_{b \to \infty} \int_1^b \frac{1}{x(x+1)}\,dx = \lim_{b \to \infty} \left. ln \left| \frac{x}{x+1} \right| \right|_1^b = \lim_{b \to \infty} ln \left| \frac{b}{b+1} \right| - \ln \frac{1}{2}$
$= - \ln \frac{1}{2} = \ln 2$

which is finite so by the integral test the series converges.

3. $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}
$

$1 = A(x+1) + Bx$

$x = 0$ ... $A = 1$

$x = -1$ ... $B = -1$

$\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$

$\lim_{b \to \infty} \int_1^b \frac{1}{x} - \frac{1}{x+1} \, dx
$

$\lim_{b \to \infty} \left[\ln{x} - \ln(x+1)\right]_1^b
$

$\lim_{b \to \infty} \left[\ln\left(\frac{x}{x+1}\right)\right]_1^b$

$\lim_{b \to \infty} \left[\ln\left(\frac{b}{b+1}\right) - \ln\left(\frac{1}{2}\right)\right]$

$\lim_{b \to \infty} \left[\ln\left(\frac{1}{1+\frac{1}{b}}\right) - \ln\left(\frac{1}{2}\right)\right] = 0 - \ln\left(\frac{1}{2}\right) = \ln(2)
$

$\frac{1}{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1},$ and that leads to danny's result.