## Sequences

1. So we tried dealing with infinite series as if they were polynomials. This caused problems.

One solution was to consider the sequence of partial sums, and treat the series converging to whatever the sequence converges to.

But we have a new problem: what’s the criteria for a sequence converging?

2. We define a “Sequence” denoted $\{a_{n}\}$ is a function whose domain is the set of all positive integers and whose range is a subset of the real numbers $\mathbb{R}$.

Example 1. Consider $a_{n}=n^{-3}$. We see as $n\to\infty$ we have $a_{n}\to 0$.

We write this as $\displaystyle\lim_{n\to\infty}\frac{1}{n^{3}}=0$.

Example 2. Consider $a_{n}=\cos(n\pi/2)$. So $a_{2k-1}=0$ for $k=1,2,3,\dots$ and $a_{2k}=(-1)^{k}$.

The sequence doesn’t settle down, it’s always oscillating. So the limit does not exist.

3. A sequence $\{a_{n}\}$ “Converges” if there is a finite number $L$ such that $\displaystyle \lim_{n\to\infty}a_{n}=L$.

A sequence “Diverges” if $\displaystyle \lim_{n\to\infty}a_{n}=\pm\infty$ or if the limit does not exist.

A sequence either converges or diverges. There is no middle ground, and there is no sequence that neither diverges nor converges. It must do one or the other.

Example 3. Consider the sequence
(1)$\displaystyle a_{n}=\frac{3n^{3}+2n+1}{7n^{3}+4}$
Does it converge or diverge? Well, one trick we can do is divide the top and bottom by $n^{3}$, giving us
(2)\displaystyle \begin{aligned}a_{n}&=\frac{3n^{3}+2n+1}{7n^{3}+4}\left(\frac{1/n^{3}}{1/n^{3}}\right)\\ &=\frac{3+2n^{-2}+n^{-3}}{7+4n^{-3}}\\ &=\frac{3+\mathcal{O}(n^{-1})}{7+\mathcal{O}(n^{-1})} \end{aligned}
Taking the limit as $n\to\infty$ becomes easier, since it amounts to setting the $\mathcal{O}(n^{-1})$ terms to zero. That is:
(3)$\displaystyle \lim_{n\to\infty}\frac{3+\mathcal{O}(n^{-1})}{7+\mathcal{O}(n^{-1})} =\frac{3+0}{7+0}=\frac{3}{7}$.
Moreover, the sequence converges.

Example 4. Consider the sequence
(5)$\displaystyle a_{n}=\frac{2n^{4}+5n+8}{n^{2}+1}$
We divide the top and bottom by $n^{2}$
(6)$\displaystyle \frac{2n^{4}+5n+8}{n^{2}+1}\left(\frac{n^{-2}}{n^{-2}}\right)=\frac{2n^{2}+5n^{-1}+8n^{-2}}{1+n^{-2}}$
Again we write this using big O notation,
(7)$\displaystyle \frac{2n^{2}+5n^{-1}+8n^{-2}}{1+n^{-2}}=\frac{2n^{2}+\mathcal{O}(n^{-1})}{1+\mathcal{O}(n^{-1})}$
Taking the limit as $n\to\infty$ produces a bit of a problem, because the sequence behaves like $2n^{2}$. How can we say this? Because the bonus parts grouped into the $\mathcal{O}(n^{-1})$ terms are effectively thrown away (they vanish when we take the limit). Whatever is left over tells us the sequence’s behaviour.

So we see that
(8)$\displaystyle \lim_{n\to\infty}\frac{2n^{2}+\mathcal{O}(n^{-1})}{1+\mathcal{O}(n^{-1})}=+\infty$
which implies the sequence from Equation (5) diverges.

4. Formal Criteria, Useful Theorems. We can be more precise. Given a sequence $\{a_{n}\}$ we have $\displaystyle\lim_{n\to\infty}a_{n}=L$ if and only if for any $\varepsilon>0$ there exists a positive integer $N$ such that $n>N$ implies $|a_{n}-L|<\varepsilon$.

Theorem 1. If $-1, then $\lim_{n\to\infty}r^{n}=0$.

Sandwich Theorem. If $a_{n}\leq b_{n}\leq c_{n}$ for each $n=1,2,3,\dots$, and
$\displaystyle \lim_{n\to\infty}a_{n}=\lim_{n\to\infty}c_{n}=L$
then $\displaystyle\lim_{n\to\infty}b_{n}=L$.

Theorem 2. If $\lim_{n\to\infty}|a_{n}|=0$, then $\lim_{n\to\infty}a_{n}=0$.

Remark. The proof is easy, since $-|a_{n}|\leq a_{n}\leq|a_{n}|$ for each $n$. Then apply the Sandwich theorem.

Theorem 3. For any constant $k$ we have
$\displaystyle \lim_{n\to\infty}\frac{k^{n}}{n!}=0$.

And we always should remember L’Hopital’s rule when considering tricky limits.