Can someone help me prove that the following integral diverges by comparison I'm kind of stuck
$\displaystyle
\int_{0}^{\infty }{\frac{dx}{\sqrt{1+x^{2}}}}
$
Thanks
Let $\displaystyle a_{n}= \frac{1}{\sqrt{1+n^2}} $ and let $\displaystyle b_{n}=\frac{1}{\sqrt{n^2}} = \frac{1}{n} $
The Limit Comparison test says that if the ratio $\displaystyle \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.
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.
We know this $\displaystyle 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 $\displaystyle \int_0^1 {\frac{{dx}}{{\sqrt {1 + x^2 } }}} $ is finite.