# 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. $$\liminf_{n\to\infty} X_n^2 > 4 \implies \limsup_{n\to\infty} X_n > 4$$

Please Help me to Prove or Disprove the given options.

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,...)

Please help me to Prove option 2.

• Have you tried anything? It'd be easier and better for you if you show what you tried and we help you with what you are having trouble. – ABP Jan 3 '20 at 2:22
• 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
• Please help me how to prove option 2. – Baljeet Jan 3 '20 at 3:13

## 1 Answer

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$$.

• Thanks for the help. – Baljeet Jan 3 '20 at 5:38
• what if we take the sequence {0,-1.0,-1,0,-1,_ _ _ _ _ _}? – Mansi Aug 19 '20 at 2:55
• @Huny Which of the four assertions are you proposing this as a counterexample? – Math1000 Aug 19 '20 at 15:32
• Sorry, didn't notice non-negative real numbers!! Thanks for replying. :) – Mansi Aug 20 '20 at 11:54