1. ## Comparison Test

$\displaystyle \int_1^\infty \frac{1}{\sqrt{\theta+1}}d\theta$

My professor tells us to create an intuition and then create a proof.

What I have so far is

$\displaystyle Let f(\theta) = \frac{1}{\sqrt{\theta+1}}d\theta$
$\displaystyle When \theta \rightarrow \infty; f(\theta)$ acts like $\displaystyle \frac{1}{\sqrt{\theta}}$
So I guess that $\displaystyle f(\theta)$ diverges because p = $\displaystyle \frac{1}{2}$

Now I need an easy, smaller function that I can prove that if that smaller function diverges, then $\displaystyle f(\theta)$ diverges, am missing something here?

2. Originally Posted by Latszer
$\displaystyle \int_1^\infty \frac{1}{\sqrt{\theta+1}}d\theta$

My professor tells us to create an intuition and then create a proof.

What I have so far is

$\displaystyle Let f(\theta) = \frac{1}{\sqrt{\theta+1}}d\theta$
$\displaystyle When \theta \rightarrow \infty; f(\theta)$ acts like $\displaystyle \frac{1}{\sqrt{\theta}}$
So I guess that $\displaystyle f(\theta)$ diverges because p = $\displaystyle \frac{1}{2}$

Now I need an easy, smaller function that I can prove that if that smaller function diverges, then $\displaystyle f(\theta)$ diverges, am missing something here?

$\displaystyle \int \dfrac{dx}{\sqrt{x+x}}= \dfrac{1}{\sqrt{2}}\int \dfrac{dx}{\sqrt{x}} < \int \dfrac{dx}{\sqrt{x+1}}$

I think now its correct

3. $\displaystyle \frac{1}{x}$ is not less than $\displaystyle \frac{1}{x+1}$ as x approaches infinity, so while your logic makes sense it technically does not "prove" the integral.

4. Is it a sufficient proof if I just sub the integral like....

$\displaystyle \int_1^\infty\frac{d\theta}{\sqrt{\theta+1}} = \lim_{c\to\infty}\int_1^c \frac{d\theta}{\sqrt{\theta+1}}$

let $\displaystyle u=\theta + 1$ ; $\displaystyle du = d\theta$
when $\displaystyle \theta = c \rightarrow u = c + 1$ ; when $\displaystyle \theta = 1 \rightarrow u = 2$
Let $\displaystyle d = c+1$
When $\displaystyle c \rightarrow\infty$ , $\displaystyle d \rightarrow\infty$

$\displaystyle \lim_{d\to\infty}\int_2^d\frac{du}{\sqrt{u}}$ = $\displaystyle \lim_{d\to\infty}\int_1^\infty\frac{du}{\sqrt{u}}d u$ - $\displaystyle \int_1^2\frac{du}{\sqrt{u}}du$

And then since $\displaystyle \lim_{d\to\infty}\int_1^d\frac{du}{\sqrt{u}}$ diverges, then $\displaystyle \int_1^\infty\frac{d\theta}{\sqrt{\theta+1}}$ diverges.

Did I do anything "illegal"? Is that a sufficient proof?

-Tyler

5. Originally Posted by Latszer
Is it a sufficient proof if I just sub the integral like....

$\displaystyle \int_1^\infty\frac{d\theta}{\sqrt{\theta+1}} = \lim_{c\to\infty}\int_1^c \frac{d\theta}{\sqrt{\theta+1}}$

let $\displaystyle u=\theta + 1$ ; $\displaystyle du = d\theta$
when $\displaystyle \theta = c \rightarrow u = c + 1$ ; when $\displaystyle \theta = 1 \rightarrow u = 2$
Let $\displaystyle d = c+1$
When $\displaystyle c \rightarrow\infty$ , $\displaystyle d \rightarrow\infty$

$\displaystyle \lim_{d\to\infty}\int_2^d\frac{du}{\sqrt{u}}$ = $\displaystyle \lim_{d\to\infty}\int_1^\infty\frac{du}{\sqrt{u}}d u$ - $\displaystyle \int_1^2\frac{du}{\sqrt{u}}du$

And then since $\displaystyle \lim_{d\to\infty}\int_1^d\frac{du}{\sqrt{u}}$ diverges, then $\displaystyle \int_1^\infty\frac{d\theta}{\sqrt{\theta+1}}$ diverges.

Did I do anything "illegal"? Is that a sufficient proof?

-Tyler
Why not just realise that $\displaystyle \frac{1}{\sqrt{\theta + 1}} = (\theta + 1)^{-\frac{1}{2}}$?

So $\displaystyle \int{\frac{1}{\sqrt{\theta + 1}}\,d\theta} = \int{(\theta + 1)^{-\frac{1}{2}}\,d\theta}$

$\displaystyle = 2(\theta + 1)^{\frac{1}{2}} + C$

$\displaystyle = 2\sqrt{\theta + 1} + C$.

So $\displaystyle \int_1^{\infty}{\frac{1}{\sqrt{\theta + 1}}\,d\theta} = \lim_{\epsilon \to \infty}\left[2\sqrt{\theta + 1}\right]_1^{\epsilon}$

$\displaystyle = \lim_{\epsilon \to \infty}2\sqrt{\epsilon + 1} - 2\sqrt{1 + 1}$

This is clearly divergent, as the square root function is always increasing.