Sequences and Their Limits

Consider the sequence $\left\{{\frac {1}{n}}\right\}_{n=1}^{\infty }=\left\{1,{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},...\right\}$. As $n\to \infty$, it appears as though ${\frac {1}{n}}\to 0$. In fact, we know that this is true since $\lim _{n\to \infty }{\frac {1}{n}}=0$. We will now formalize the definition of a limit with regards to sequences.

Definition: If $\{a_{n}\}$ is a sequence, then $\lim _{n\to \infty }a_{n}=L$ means that for every $\epsilon >0$ there exists a corresponding $N\in \mathbb {N}$ such that if $n\geq N$, then $\mid a_{n}-L\mid <\epsilon$. If this limit exists, then we say that the sequence $\{a_{n}\}$ Converges, and if this limit doesn't exist then we say say the sequence $\{a_{n}\}$ Diverges.

We note that our definition of the limit of a sequence is very similar to the limit of a function, in fact, we can think of a sequence as a function whose domain is the set of natural numbers $\mathbb {N}$. From this notion, we obtain the very important theorem:

Theorem 1: If $\{a_{n}\}$ is a sequence and a function $f(x)$, then if $f(n)=a_{n}$ where $n\in \mathbb {N}$ and $\lim _{x\to \infty }f(x)=L$, then $\lim _{n\to \infty }f(n)=L$.

• Proof of Theorem 1: Let $\epsilon >0$ be given. We know that $\lim _{x\to \infty }f(x)=L$ which implies that $\forall \epsilon >0\exists N\in \mathbb {R}$ such that if $x\geq k$ then $\mid f(x)-L\mid <\epsilon$.

• Now we want to show that $\forall \epsilon >0\exists N\in \mathbb {N}$ such that if $n\geq N$ then $\mid a_{n}-L\mid <\epsilon$. We will choose $N=\mathrm {max} \{1,\lceil k\rceil \}$. This ensures that $N$ is an integer.

• Now since $N\geq k$ then it follows that if $\forall n\geq N$ then $\mid f(n)-L\mid <\epsilon$. But $f(n)=a_{n}$ and so $\mid a_{n}-L\mid <\epsilon$ so $\lim _{n\to \infty }a_{n}=L$. $\blacksquare$

Important Note: The converse of this theorem is not implied to be true! That is if $f(n)=a_{n}$ and $\lim _{n\to \infty }f(n)=L$, then this does NOT imply that $\lim _{x\to \infty }f(x)=L$. For example, consider the sequence $\{\sin(2\pi n)\}=\{0,0,0,...\}$. Clearly this sequence converges at 0. However, the function $f(x)=\sin(2\pi x)$ does not converge, instead, it diverges as it oscillates between $-1$ and $1$.

For example, consider the sequence $\left\{{\frac {n+2}{n}}\right\}_{n=1}^{\infty }$. If we let $f(n)={\frac {n+2}{n}}$ be a function whose domain is the natural numbers, then we calculate the limit of this function like we have in the past, namely:

${\begin{aligned}\lim _{n\to \infty }f(n)=\lim _{n\to \infty }{\frac {n+2}{n}}=\lim _{n\to \infty }1+{\frac {2}{n}}=1\end{aligned}}$

Therefore the limit of our sequence $f(n)=a_{n}$ is 1, that is, $\left\{{\frac {n+2}{n}}\right\}_{n=1}^{\infty }$ converges to 1 as $n\to \infty$.

Now let's look at another major theorem.

Theorem 2: If $\lim _{n\to \infty }a_{n}=L$ and a function $f$ is continuous at $L$, then $\lim _{n\to \infty }f(a_{n})=f\left(\lim _{n\to \infty }a_{n}\right)=f(L)$.

• Proof of Theorem 2: If $f$ is a continuous function at $x=L$, then we know that $f(L)$ is defined and that $\lim _{x\to L}f(x)=f(L)$. By the definition of a limit, $\forall \epsilon >0\exists \delta >0$ such that if $0<\mid x-L\mid <\delta$ then $\mid f(x)-f(L)\mid <\epsilon$. Let $x=a_{n}$ so then if $0<\mid a_{n}-L\mid <\delta$ then $\mid f(a_{n})-f(L)\mid <\epsilon$.

• We want to show that $\lim _{n\to \infty }f(a_{n})=f(L)$, that is $\forall \epsilon >0\exists N\in \mathbb {N}$ such that if $n\geq N$ then $\mid f(a_{n})-f(L)\mid <\epsilon$, which is what we showed above.

We will now look at some important limit laws regarding sequences