Is the integral of $\frac{1}{x(x+1)}$ divergent?

Question

$$ \,(Set\,up)\int_1^{\infty} \frac{dx}{x(x+1)} = \lim_{n\to \infty} \int_1^{n} \frac{dx}{x(x+1)}$$

$$ \, (Integrated)\lim_{n\to \infty} \int_1^{n} \frac{1}{x} + \frac{-1}{x+1} $$ $$ \,(Evaluate)\lim_{n\to \infty} [ln{\lvert x\rvert-ln\lvert x+1\rvert]} |_1^{n}$$

After doing some simplification: $$ = \frac{ln\lvert x\rvert}{ln\lvert x+1|} + ln2 $$

If I were to say the denominator grows faster, then the integral would come out to be $0$, but whether or not I would consider that additional $+1$ significant, I do not know.

I will conclude that this integral is convergent, but I would like to know if I am correct or not. I would like to understand this a little more. I have not delved into series expansion yet. I watched a Khan Academy video earlier, and he hit on a similar point with limit comparisons (or something along those lines) to explain a problem like this.

Bring anything you know (supplemental or not).

Thanks, Michael

Answer 1

It's $$\int\limits_1^{+\infty}\frac{1}{x(x+1)}dx=\int\limits_1^{+\infty}\left(\frac{1}{x}-\frac{1}{x+1}\right)dx=\left(\ln{|x|}-\ln(|x+1|\right)_1^{+\infty}=\ln\frac{x}{x+1}|_1^{+\infty}=$$ $$=\lim_{x\rightarrow+\infty}\ln\frac{x}{x+1}-\ln\frac{1}{2}=\lim_{x\rightarrow+\infty}\ln\frac{1}{1+\frac{1}{x}}+\ln2=\ln2.$$ Because $-\ln\frac{1}{2}=-\ln2^{-1}=-(-1)\ln2=\ln2.$

Answer 2

Since for $x\ge 1,$ 

$$0<\frac{1}{x(1+x)}<\frac{1}{x^2}\,\,\text { and } \int_1^\infty \frac{1}{x^2}\,dx <\infty,$$

your integral converges by the comparison test.