1. ## Integral Convergence

Well, I'm stuck with only one problem left on my homework, and I'm getting somewhat frustrated at it. We're trying to determine if an integral converges. It looks something like this:

http://i141.photobucket.com/albums/r...onvProblem.jpg

I'm almost certain it's convergence can be determined through a direct comparison test, but I can't think of a function that could help.

Anyone have any idea?

2. I've not been doing this for a few time, but what about:

The function is positive and continuous on $[1,+\infty[$

$\forall x>1,\ 0< ln(x) < x$ and $0<\sqrt{x} < x$, so $\frac{1}{\sqrt{x}+ln(x)} > \frac{1}{2x}$.
So the integral is divergent.

3. Originally Posted by EightballLock
Well, I'm stuck with only one problem left on my homework, and I'm getting somewhat frustrated at it. We're trying to determine if an integral converges. It looks something like this:

http://i141.photobucket.com/albums/r...onvProblem.jpg

I'm almost certain it's convergence can be determined through a direct comparison test, but I can't think of a function that could help.

Anyone have any idea?
I have a million ideas. I will present only one that I enjoy. Consider this:

Let $f(x)=\frac{1}{\sqrt{x}+\ln(x)}$. Then it is apparent that $\forall{x}\in[1,\infty)~f(x)>0\wedge{f(x)\in\mathcal{C}}\wedge{f (x)\in\downarrow}$

Therefore this integral shares convergence/ divergence with $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}+\ln(n)}$

Now we can see that if we define $b_n=\frac{1}{\sqrt{n}}$ that $\lim_{n\to\infty}\frac{a_n}{b_n}=1\implies\sum{a_n }\text{ and }\sum{b_n}\text{ share divergence}$. Thus your integral is divergent.

Note that there is a limit comparison test analogue with integrals, but it is not often taught so I used this instead.

4. Oh wow, I hadn't considered that one. But yeah, integrating it yields $Ln(x)/2 + C$, which diverges as x approaches infinity. And yeah, it's less than the original function over the interval, so the original function must also diverge.

....Wow, that was easier than I thought.

Thank you so much man; I'm really impressed at this site & community and how helpful it is.'

EDIT: The second post must have popped up while I was writing this post.
Originally Posted by Mathstud28
Note that there is a limit comparison test analogue with integrals, but it is not often taught so I used this instead.
Yeah, we've been taught it, but I can't think of a function that you could divide this function by (or be divided by this function) that would yield a positive constant, which is apparently one of requisites for limit comparison testing.

5. Originally Posted by EightballLock
Oh wow, I hadn't considered that one. But yeah, integrating it yields $Ln(x)/2 + C$, which diverges as x approaches infinity. And yeah, it's less than the original function over the interval, so the original function must also diverge.

....Wow, that was easier than I thought.

Thank you so much man; I'm really impressed at this site & community and how helpful it is.
People here are always glad to help! FYI \ln(x) gives $\ln(x)$ and \frac{whatever}{you\cdot{want}} gives $\frac{whatever}{you\cdot{want}}$