How to find the limit of $\frac{\ln(n+1)}{\sqrt{n}}$ as $n\to\infty$? I'm working on finding whether sequences converge or diverge. If it converges, I need to find where it converges to.
From my understanding, to find whether a sequence converges, I simply have to find the limit of the function.
I'm having trouble getting started on this one (as well as one more, but I'll stick to one at a time).
I would appreciate if someone could explain how I should start this one.

You need to use L'Hopital's rule, twice. What this rule states is that basically you if you are trying to take the limit of something that looks like $\frac{f(x)}{g(x)}$, you can keep on taking the derivative of the numerator and denominator until you get a simple form where the limit is obvious. Note: L'Hopitals rule doesn't always work.
We have: 
$$\lim_{n\to \infty} \frac{ln(n+1)}{\sqrt{n}}  $$
We can keep on taking the derivative of the numerator and denominator to simplify this into a form where the limit is obvious:
$$= \lim_{n\to \infty}  \frac{\frac{1}{n+1}}{\frac{1}{2\sqrt{n}}} $$
$$= \lim_{n\to \infty}  \frac{(n+1)^{-1}}{(2\sqrt{n})^{-1}} $$
$$= \lim_{n\to \infty}  \frac{2\sqrt{n}}{n+1} $$
Hrm - still not clear. Let's apply L'Hopitals rule one more time:
$$= \lim_{n\to \infty}  \frac{\frac{1}{\sqrt{n}}}{1} $$
$$= \lim_{n\to \infty}  \frac{1}{\sqrt{n}} $$
The limit should now be obvious. It is now of the form $\frac{1}{\infty}$, which equals zero.
$$\lim_{n\to \infty} \frac{ln(n+1)}{\sqrt{n}} = 0  $$
