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.
 A: 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.
$$\lim_{n\to \infty} \frac{ln(n+1)}{\sqrt{n}} = 0  $$
A: We use that $$\log x=\int_1^x t^{-1}dt$$
Let $\alpha>0$; choose $0<\varepsilon <\alpha$. Then $$\frac{\log x}{x^{\alpha}}=\frac{1}{x^{\alpha}}\int_1^x t^{-1}dt<\frac{1}{x^{\alpha}}\int_1^x t^{\varepsilon-1}dt<\frac{x^{\varepsilon-\alpha}}{\varepsilon}\to 0$$
A: I assume you want the limit as $n \to \infty$.  $\ln n$ grows more slowly than any positive power of $n$ as $n \to \infty$, including $\sqrt{n}$.  The $+1$ in $\ln(n+1)$ is basically a distraction.  So the answer is zero.
This is a little like killing a mosquito with a sledgehammer, but the conditions of L'Hopital's Rule are met (check them), you can apply it, and it works out nicely.
A: Use L'Hospital's rule.  Namely, if $\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}=L$,
then $\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=L$.
In your case just take in terms of x rather than n, so $f(x)=\ln(x+1)$ and $g(x)=\sqrt(x)$, then take the derivatives and find the limit.
A: Let's consider the subsequence of the sequence $a_n = \frac{\ln(n+1)}{\sqrt{n}}$ where $n = e^m-1$.  This subsequence is given by
$$b_m = \frac{m}{\sqrt{e^m-1}}$$
We have that
$$0 < b_m < \frac{m}{\sqrt{e^{m-1}}} \qquad\mbox{for } m \ge 2$$
and as $m\to\infty$, the right-hand side of this inequality goes to zero, so $(b_m)$ converges to $0$ by the Squeeze Theorem.  
Of course, having a subsequence which converges does not mean a sequence converges in general, but it does for sequences which are eventually monotone increasing/decreasing.  We'll show that $a_n$ is eventually monotone decreasing.  
Let
$$f(x) = \frac{\ln(x+1)}{\sqrt{x}}$$
Then, 
$$f'(x) = \frac{\sqrt{x}\frac{1}{x+1}-\frac{1}{2\sqrt{x}}\ln(x+1)}{x} = \frac{2x - (x+1)\ln(x+1)}{2x^{\frac{3}{2}}(x+1)}$$
and for $x$ sufficiently large, say, $x>X$, $(x+1)\ln(x+1) > 2x$ (since $\ln(x+1)$ certainly grows larger than $2$ for $x$ large!), so the derivative will be negative for $x>X$ and $a_n$ will decrease for $n>X$.  Since we've shown that a subsequence of $a_n$ converges, $a_n$ also converges to the same limit.
