# Math Help - Divergence by comparison

1. ## Divergence by comparison

Can someone help me prove that the following integral diverges by comparison I'm kind of stuck

$
\int_{0}^{\infty }{\frac{dx}{\sqrt{1+x^{2}}}}
$

Thanks

2. Originally Posted by jarny
Can someone help me prove that the following integral diverges by comparison I'm kind of stuck

$
\int_{0}^{\infty }{\frac{dx}{\sqrt{1+x^{2}}}}
$

Thanks
Hi

$\frac{1}{\sqrt{1+x^2}}$ is equivalent to $\frac{1}{x}$ in $+\infty$

3. That wont work. I know that but that integral is mathematically greater than my integral and a comparison won't work

4. Originally Posted by running-gag
Hi

$\frac{1}{\sqrt{1+x^2}}$ is equivalent to $\frac{1}{x}$ in $+\infty$

Let $a_{n}= \frac{1}{\sqrt{1+n^2}}$ and let $b_{n}=\frac{1}{\sqrt{n^2}} = \frac{1}{n}$

The Limit Comparison test says that if the ratio $\frac{a_{n}}{b_{n}}$ tends to a finite positive number (such as 1?), then the two series behave the same way. The Harmonic diverges, so your series must diverge as well.

I hope this helps.

5. Not really I need something in the form

1+x^2 < _____

so

1/sqrt(1+x^2) > __________

and S __________ diverges so by comparison S1/sqrt(1+x^2) must diverge

6. Hmm... I understand what you are asking, but I can't think of a candidate for the Comparison Test to be applied because the Harmonic Series is 'above' the given series, so comparing term by term does not help conclude that what you have diverges. The Limit Comparison Test, however, helps answer this question as I have done above.

7. We know this $x \geqslant 1 \Rightarrow \quad \frac{1}
{{\sqrt {1 + x^2 } }} \geqslant \frac{1}
{{\sqrt 2 x}} \Rightarrow \quad \int_1^\infty {\frac{{dx}}
{{\sqrt {1 + x^2 } }} \geqslant \int_1^\infty {\frac{{dx}}
{{\sqrt 2 x}}} }$

We also know that $\int_0^1 {\frac{{dx}}{{\sqrt {1 + x^2 } }}}$ is finite.

8. Plato 1/ sqrt(2x) is greater than 1/sqrt( 1+ x^2)

I think you screwed up

9. Actually I think I got it

10. Originally Posted by jarny
Plato 1/ sqrt(2x) is greater than 1/sqrt( 1+ x^2)
I think you screwed up
It is $\frac{1}{{\left( {\sqrt 2 } \right)x}}$.