Q:

Limit of a sequence bounded by a convergent sequence

Suppose I have a sequence of real numbers $\{X_n\}$ such that $|X_n|<\frac{1}{n}$. I want to show that $\lim_{n \rightarrow \infty }X_n=0$. Any suggestion?

A:

A standard epsilon based convergence argument will work for you here. Don't think of it as being bounded by a convergent sequence. Rather, you've been directly given the fact that the distance between $x_n$ and $0$ goes to $0$ as $n \to \infty$.
To formalize this argument: a sequence $x_n \in \mathbb{R}$ converges to a value $x$ if, for all $\epsilon > 0$ there exists $N \in \mathbb{Z}$ such that if $n > N$, then $\left| x_n - x \right| < \epsilon$. In this case, for any $\epsilon > 0$ there exists $N$ sufficiently large such that $\frac{1}{N} < \epsilon$. And if $n > N$, then $$\left| x_n - 0 \right| = \left| x_n \right| < \frac{1}{n} < \frac{1}{N}.$$

