# Bounded Sequences and Extrapolation of Convergence From Related Sequences

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.

• I am curious. This is the second post I have seen of yours within 2 days. I'm interested in knowing what you are studying for and why you are concerned with the level of rigor. Of course, you do not need to reply. I am just interested.
– Eoin
Commented Sep 17, 2014 at 3:49
• I am attempting to develop a foundation in elementary pure mathematics (i.e. Analysis and Abstract Algebra (just linear, now actually). I'm so concerned with the rigor because my main problems in my attempts at proofs are subconcious extensions of basic mathematical axioms beyond what the axiom is actually expressing. This leads to an oversimplification of my proofs, to the point that they're not proofs. By pushing for more rigor, I'm trying to stem this simplification. In so doing, the math, I believe, becomes more understandable as well. That was quite long, sorry.
– TQM
Commented Sep 19, 2014 at 9:18
• No no. I understand. In much the same way I have been trying to do the same! I have troubles proving something when I believe it is intuitively obvious. We are in the same boat! And I believe that you are doing a good job and appreciate that you are going in this direction. There is a quote somewhere about someone saying "I have never seen a complete proof of someone but I have never seen an incorrect one either". It's just a characteristic of modern math that we are concerned with absolute completeness. Anyways good job!
– Eoin
Commented Sep 19, 2014 at 16:54

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,

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

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.

• Didn't even occur to me to prove this way. Thanks you for the unique insight!
– TQM
Commented Sep 19, 2014 at 9:34

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 $\frac{S_n}{n} \to 0$.