obtaining inequality for the DCT

Question

I'm trying to solve the integral using specifically the direct comparison test $$\int_{1}^{\infty}\frac{\sqrt{x+1}}{x^2}dx$$ Now i know that $\frac{1}{x^2}\leq\frac{\sqrt{x+1}}{x^2}$. I obtained this by graphing the functions online. However, is there any way i could have obtained this inequility without looking at the graph?

Answer 1

For any $x\geq 0$, it is clear that $\sqrt{1+x}\geq 1$. You can see this by taking square of both sides. Also $x^2\geq 0$ for any $x$. For the interval $[1,\infty]$, it is clear that $x^2>0$. So if you combine all these info you get

$$ \frac{1}{x^2}\leq \frac{\sqrt{1+x}}{x^2} $$

Answer 2

Note that, for $x\ge1$, $$ \frac{\sqrt{1+x}}{x^2}\le\frac{\sqrt{2x}}{x^2}=\frac{\sqrt2}{x^{3/2}}. $$ Since $\int_1^\infty\frac{1}{x^{3/2}}dx$ converges, so does $\int_1^\infty\frac{\sqrt{1+x}}{x^2}dx$.

Answer 3

For large $x$, $\sqrt{x+1}$ behaves like $x^{1/2}$ ($1$ becomes neglectible) and your integrand like $x^{-3/2}$ hence the integral converges.

If you really want an upper bound, there will certainly be an $x_0$ and a $c>1$ such that

$$x>x_0\implies c\sqrt x>\sqrt{x+1}.$$

Indeed, after rewriting we have

$$(c^2-1)x>1$$ for $$x>x_0=\frac1{c^2-1}.$$

(@xpaul took $c=\sqrt2$.)