Sequence $\ln(n)/n$ to $0$ 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 $n$.
 A: 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.
A: 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 $n\gt4/\epsilon^2$.
A: To find the maximum of a sequence consider it as a function, namely
$$f(x) = \frac{\ln x}{ x}.$$
Assuming $x>0$ we find the extremum of this function
$$f'(x) = \frac{1-\ln x}{x^2}=0 \Rightarrow \ln x =1 \Rightarrow x=e.$$
Then
$$\sup_{x \in [1,\infty)} \frac{\ln x}{x} \leq \frac{1}{e} \Rightarrow | \frac{\ln n}{n}| \leq \frac{1}{e} \quad \forall n > 0 .$$
The second sequence is clear since we have
$$|\frac{1-n}{n}| = |1- \frac{1}{n}| < 1 \quad \forall n > 0$$
