# Proving $\limsup x_n=0\implies\lim x_n^2=0$

Let $$X_n$$ be a sequence of non negative real numbers. Then which of the following is true?

1. $$\liminf_{n\to\infty} X_n = 0 \implies \lim_{n\to\infty} X_n^2 = 0$$
2. $$\limsup_{n\to\infty} X_n = 0 \implies \lim_{n\to\infty} X_n^2 = 0$$
3. $$\liminf_{n\to\infty} X_n = 0 \implies X_n$$ is bounded
4. $$\liminf_{n\to\infty} X_n^2 > 4 \implies \limsup_{n\to\infty} X_n > 4$$

My Attempt:

Option 1 can be discarded by taking the sequence (0,1,0,1,0,1,0,1,...)

Option 3 can be discarded by taking the sequence (0,1,0,2,0,3,0,4,0,5,0,6,...)

Option 4 can be discarded by taking the sequence (3,3,3,3,3,...)

• Think about what $\limsup_{n\to\infty}x_n=0$ means for a sequence of non-negative numbers. – bjorn93 Jan 3 '20 at 2:23
If $$\limsup_{n\to\infty}x_n=0$$, then there exists $$N$$ such that $$\sup_{n\geqslant N}x_n<1$$, and hence $$x_n^2 for $$n\geqslant N$$. It follows then $$\limsup_{n\to\infty} x_n^2\leqslant \limsup_{n\to\infty} x_n=0,$$ and from $$x_n\geqslant 0$$ that $$\liminf_{n\to\infty} x_n^2\geqslant 0$$. Since trivially $$\liminf_{n\to\infty} x_n^2\leqslant\limsup_{n\to\infty} x_n^2$$, we find that $$\liminf_{n\to\infty}x_n^2 = \limsup_{n\to\infty} x_n^2 = 0,$$ and hence $$\lim_{n\to\infty} x_n^2=0$$.