I came across the following problems on null sequences during the course of my self-study of real analysis.
Let $x_n = \sqrt{n+1}- \sqrt{n}$. Is $(x_n)$ a null sequence?
Consider $y_n = \sqrt{n+1}+ \sqrt{n}$. Then $x_{n}y_{n} = 1$ for all $n$. So either $(x_n)$ or $(y_n)$ is not a null sequence. It seems $(y_n)$ is not a null sequence. I think $(x_n)$ is a null sequence because $\sqrt{n+1} \approx \sqrt{n}$ for large $n$ which implies that $x_n \approx 0$ for large $n$.
If $(x_n)$ is a null sequence and $y_n = (x_1+ x_2+ \dots + x_n)/n$ then $(y_n)$ is a null sequence.
Suppose $|x_n| \leq \epsilon$ for all $n >N$. If $n>N$, then $y_n = y_{N}(N/n)+ (x_{N+1}+ \dots+ x_n)/n$. From here what should I do?
If $p: \mathbb{R} \to \mathbb{R}$ is a polynomial function without constant term and $(x_n)$ is a null sequence, then $p((x_n))$ is null.
We know that $|x_n| \leq \epsilon$ for all $n>N$. We want to show that $|p(x_n)| \leq \epsilon$ for all $n>N_1$. We know that $p(x) = a_{d}x^{d} + \cdots+ a_{1}x$. So $|p(x_n)| \leq a_{d} \epsilon^{d} + \cdots+ a_{1} \epsilon$
for all $n>N_1$.