Comparison Test

**Latszer** · February 16th 2010, 05:25 PM

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

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

What I have so far is

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

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

**felper** · February 16th 2010, 06:14 PM

Originally Posted by Latszer

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

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

What I have so far is

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

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

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

I think now its correct

**Latszer** · February 16th 2010, 06:21 PM

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

**Latszer** · February 16th 2010, 07:25 PM

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

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

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

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

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

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

-Tyler

**Prove It** · February 16th 2010, 07:35 PM

Originally Posted by Latszer

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

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

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

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

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

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

-Tyler

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

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

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

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

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

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

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

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