As the title says, I'm trying to prove that the sequence $$a_n=\dfrac{\ln(n+7)}{n}$$ converges. I'm trying to use the squeeze theorem and l'hopitals rule, but I'm not making any progress. I just ran a quick simulation and it seems that $a_n$ converges to $0$.
How do I go about proving this?
First write: $\ln(n+7) = \ln n + \ln\left(1+\frac{7}{n}\right)$, and use the popular inequality: $\ln(1+x) \le x$ with $x = \frac{7}{n}$ we have:
$0 < a_n \le \dfrac{\ln(n)}{n}+\dfrac{7}{n^2}=\ln\left(\sqrt[n]{n}\right)+\dfrac{7}{n^2}\to \ln(1) + 0 = 0$. You need to verifty that $\sqrt[n]{n} \to 1$ by using a sleek AM-GM inequality as this: $1\le \sqrt[n]{n} = \sqrt[n]{1\cdot 1\cdots\sqrt{n}\cdot \sqrt{n}}\le \dfrac{1+1+\cdots + 1+ \sqrt{n}+\sqrt{n}}{n}= \dfrac{n-2+2\sqrt{n}}{n}= 1-\dfrac{2}{n}+\dfrac{2}{\sqrt{n}}$ which converges to $1$ by squeeze lemma, and the main statement follows. In doing this "proof", we avoided bothering the L'hospitale rule as much as we could.
The squeezing theorem (assume $x>0$):
$0 \leq \frac{\ln{(x+7)}}{x} \leq \frac{x+7}{x}$
It's fairly easy to see that $\frac{x+7}{x}$ converges to one, so this doesn't work.
L'Hôpital's rule says:
$\lim_{x \to \infty} \frac{\ln(x+7)}{x} = \lim_{x \to \infty} \frac{ \frac{d}{dx} \ln{(x+7)} }{ \frac{d}{dx} x } = \lim_{x \to \infty} \frac{1}{x+7} = 0$
I showed in THIS ANSWER, that the logarithm function satisfies the inequality
$$\log(n)\le \sqrt{n} \tag 1$$
for $n\ge 1$.
Using $(1)$ reveals that
$$0\le \frac{\log(n+7)}{n}\le \frac{\sqrt{n+7}}{n}\tag 2$$
whereupon applying the squeeze theorem to $(2)$ yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{\log(n+7)}{n}=0}$$
