Sequence $\ln(n)/n$ to $0$ [closed]

I'm taking a real analysis course for my second year and I'm still new when it comes to convergence of a sequence using the M-epsilon definition.

I've simplified my sequence up to $$\frac{\ln(n)}{n}<\epsilon,$$

but where do I go from here? How do I find an upper bound for $\ln(n)/n$, I have tried using $\epsilon/n$ and $(n-1)/n$ ,but it doesn't satisfy the inequality for a large $n$.

With the help of the hint in the first comment of @Gary we will solve that. We know that $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},$$ so we have that $$n\leq\sum_{k=0}^{\infty}\frac{(\sqrt{2n})^k}{k!}=e^{\sqrt{2n}}$$ which implies that $$n\leq e^{\sqrt{2n}}$$ so $$\ln(n)\leq\sqrt{2n}$$ and $$\frac{\ln(n)}{n}\leq\frac{\sqrt{2n}}{n}=\sqrt{\frac{2}{n}}.$$ We want $\sqrt{\frac{2}{n}}<\epsilon$,$\forall n\in\mathbb{N}$ which are greater than some $n_0\in\mathbb{N}$.
So $$\sqrt{\frac{2}{n}}<\epsilon$$ $$\iff\frac{2}{n}<\epsilon^2$$ $$\iff n>\frac{2}{\epsilon^2}.$$ If we choose $n_0=\frac{2}{\epsilon^2}+1$ we have that $\forall\epsilon>0$, $\exists n_0\in\mathbb{N}$ such that $\forall n>n_0$, $\frac{\ln(n)}{n}<\epsilon$, hence our sequence converges to zero by definition.

Let's use the integral definition of the natural logarithm. Note that for $n\gt1$ we have

$$\ln n=\int_1^n{dt\over t}=\int_1^\sqrt n{dt\over t}+\int_\sqrt n^n{dt\over t}\lt\int_1^\sqrt n dt+\int_\sqrt n^n{dt\over\sqrt n}\lt\int_0^\sqrt n dt+\int_0^n{dt\over\sqrt n}=\sqrt n+\sqrt n$$

Hence

$$0\lt{\ln n\over n}\lt{2\sqrt n\over n}={2\over\sqrt n}$$

so $\left|\ln n\over n\right|\lt\epsilon$ for all $n\gt4/\epsilon^2$.