# Convergence in probability and almost sure convergence

Let $$X_i \sim U[0,1]$$ are independent. Let $$Y_n=\max\{ X_1,X_2,...,X_n\}$$. And let $$Z_n=\frac{Y_n}{1+nY_n}$$.

Does $$Z_n$$ converge in probability to $$0$$?

My attempt: $$P(\frac{Y_n}{1+nY_n}\le \epsilon)=P(Y_n \le \epsilon(1+nY_n))=P(Y_n(1-\epsilon \cdot n)\le \epsilon)=...$$

If $$\epsilon$$ is very large the..

$$...=P(Y_n \ge \frac{\epsilon}{1-\epsilon \cdot n})$$

And $$\lim_{n --> \infty}P(Y_n \ge \frac{\epsilon}{1-\epsilon \cdot n})=1$$

But If $$\epsilon$$ is very small then:

$$...=P(Y_n \le \frac{\epsilon}{1-\epsilon \cdot n})$$

And $$\lim_{n --> \infty}P(Y_n \le \frac{\epsilon}{1-\epsilon \cdot n})=0$$

So, $$Z_n$$ is not converge in probability.

What with almost sure convergence?

• If something doesn't converge in probability then it doesn't converge almost surely, as a.s convergence implies convergence in probability. Jan 23 '20 at 22:18

For any $$y\in [0,1]$$, $$0\le\frac{y}{1+ny}\le \frac{1}{1+n}.$$