Bounded Sequences and Extrapolation of Convergence From Related Sequences

Question

I'm considering some sequence $S_n$ which is bounded, and I want to prove that $S_n/n$ is convergent. I'm thinking that I could simply take $lim_{n \to \infty} S_n/n$ and simplify this to $(lim_{n \to \infty} S_n)/\infty$. Then this would be convergent for any bounded sequence $S_n$. What I'm struggling to understand and thus in turn, struggling to show with more rigor, is why the property of $S_n$ being bounded results in the aforementioned simplified limit, being convergent. I feel like I may be making a wrong assumption here, and/or that I have taken the wrong path in attempting to prove such a statement.

Answer 1

Suppose $(S_n)$ is bounded. Then there is a positive quantity $M$ such that $ |S_n| \le M $.

Then let $\epsilon \gt 0$ be arbitrary. There is $m \in \Bbb N$ such that $m \gt \dfrac{M}{\epsilon}$, that is $ \epsilon \gt \dfrac M m $. Then,

$$ n \ge m \implies \epsilon \gt \dfrac M m \ge \dfrac M n \ge \dfrac{|S_n|}{n} = \left|{\dfrac{S_n}{n} - 0}\right| \implies \lim_{n} S_n = 0 $$

Answer 2

Proof. Suppose the sequence $\langle S_n\rangle_{n=1}^\infty $ is bounded. Then $\exists M\in \mathbb{R}$ such that $\forall i\in \mathbb{N}$, $S_i\leq M$. Similarly $\exists m\in \mathbb{R}$ such that $\forall i\in \mathbb{N}$, $S_i\geq m$.We continue by considering the new sequence $\langle \frac{S_n}{n} \rangle_{n=1}^\infty$.

Let $\epsilon>0$ be arbitrary. Suppose $M\geq 0$, (if not, then we consider $-M$ in the following). Since $\mathbb{N}$ is unbounded, we may find $n_1\in \mathbb{N}$ such that $n_1> \frac{M}{\epsilon/2}$. This implies $\epsilon/2>\frac{M}{n_1}$. We need not consider $m>0$ if $M\geq 0$. However, if $m<0$ and $M<-m$, or if $M<0$, then we may substitute the above with $-m$.

So for any $s_i\in \langle S_n\rangle_{n=1}^\infty$ we have $a,b\geq n_1$ implies $\frac{s_a}{a}<\epsilon/2$ and $\frac{s_b}{b}<\epsilon/2$. Then, by the triangle inequality, we have $|s_a/a-s_b/b|\leq|s_a/a|+|s_b/b|<\epsilon$. Therefore, this sequence is a Cauchy sequence and thus has a limit.

Answer 3

If by $S_n$ being bounded, you mean that there exists $M \in \mathbb{R}$ such that $|S_n| \leq M$ for each $M$, then you can use the following strategy. We can show that $\displaystyle\lim_{n \to \infty} \frac{S_n}{n} = 0$. To do this, we need to show that, for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $\forall n > N$, $|\frac{S_n}{n} - 0| = |\frac{S_n}{n}| < \epsilon$. We can observe that $$\bigg|\frac{S_n}{n}\bigg| = \frac{|S_n|}{|n|} \leq \frac{M}{n}$$ Then, given any $\epsilon > 0$, we can choose $N = \frac{M}{\epsilon}$ so that, for each $n > N$, $$n > N = \frac{M}{\epsilon} \Longrightarrow \frac{M}{n} < \epsilon \Longrightarrow \bigg|\frac{S_n}{n}\bigg| < \epsilon $$ and we have shown that $\frac{S_n}{n} \to 0$.