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

Limit Laws of Convergent Sequences

We will now look at some very important limit laws regarding convergent sequences, all of which are analogous to the limit laws for functions that we already know of.

• Law 1 (Addition Law of Convergent Sequences): If the limits of the sequences $\{a_{n}\}$ and $\{b_{n}\}$ are convergent, that is $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$ then $\lim _{n\to \infty }(a_{n}+b_{n})=\lim _{n\to \infty }a_{n}+\lim _{n\to \infty }b_{n}=A+B$.

• Proof of Law 1: Let $\{a_{n}\}$ and $\{b_{n}\}$ be convergent sequences. By the definition of a sequence being convergent, we know that $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$ for some $A,B\in \mathbb {R}$.

• Now let $\epsilon >0$ be given, and recall that $\lim _{n\to \infty }a_{n}=A$ implies that $\forall \epsilon \exists N_{1}\in \mathbb {N}$ such that if $n\geq N_{1}$ then $\mid a_{n}-A\mid <{\frac {\epsilon }{2}}$. Similarly, $\lim _{n\to \infty }b_{n}=B$ implies that $\forall \epsilon \exists N_{2}\in \mathbb {N}$ such that if $n\geq N_{2}$ then $\mid b_{n}-B\mid <{\frac {\epsilon }{2}}$.

• We will choose $N=\mathrm {max} \{N_{1},N_{2}\}$. By choosing the larger of $N_{1}$ and $N_{2}$, we ensure that the if $n\geq \mathrm {max} \{N_{1},N_{2}\}$, then $\mid a_{n}-A\mid <{\frac {\epsilon }{2}}$ and $\mid b_{n}-B\mid <{\frac {\epsilon }{2}}$. We now want to show that if $n\geq N$, then $\mid (a_{n}+b_{n})-(A+B)\mid <\epsilon$. By the triangle inequality we obtain that:

${\begin{aligned}\quad \quad \mid (a_{n}+b_{n})-(A+B)\mid =\mid (a_{n}-A)+(b_{n}-B)\mid \leq \mid a_{n}+A\mid +\mid b_{n}+B\mid <{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

• Therefore $\lim _{n\to \infty }(a_{n}+b_{n})=A+B$. $\blacksquare$

• Law 2 (Difference Law of Convergent Sequences): If the limits of the sequences $\{a_{n}\}$ and $\{b_{n}\}$ are convergent, that is $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$ then $\lim _{n\to \infty }(a_{n}-b_{n})=\lim _{n\to \infty }a_{n}-\lim _{n\to \infty }b_{n}=A-B$.

• Law 3 (Product Law of Convergent Sequences): If the limits of the sequences $\{a_{n}\}$ and $\{b_{n}\}$ are convergent, that is $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$, then $\lim _{n\to \infty }(a_{n}b_{n})=\lim _{n\to \infty }a_{n}\cdot \lim _{n\to \infty }b_{n}=AB$.

• Law 4 (Quotient Law of Convergent Sequences): If the limits of the sequences $\{a_{n}\}$ and $\{b_{n}\}$ are convergent, that is $\lim _{n\to \infty }a_{n}=A$ and $\lim _{n\to \infty }b_{n}=B$, then $\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}={\frac {\lim _{n\to \infty }a_{n}}{\lim _{n\to \infty }b_{n}}}={\frac {A}{B}}$ provided that $\lim _{n\to \infty }b_{n}\neq 0$.

• Law 5 (Constant Multiple Law of Convergent Sequences): If the limit of the sequence $\{a_{n}\}$ is convergent, that is $\lim _{n\to \infty }a_{n}=A$, and $k$ is a constant, then $\lim _{n\to \infty }ka_{n}=k\lim _{n\to \infty }a_{n}=kA$.

• Law 6 (Power Law of Convergent Sequences): If the limit of the sequence $\{a_{n}\}$ is convergent, that is $\lim _{n\to \infty }a_{n}=A$ and $k$ is a non-negative integer, then $\lim _{n\to \infty }(a_{n})^{k}=\left(\lim _{n\to \infty }a_{n}\right)^{k}=(A)^{k}$ provided that $k\in \mathbb {N}$.

• Law 7 (Squeeze Theorem for Convergent Sequences): If the limits of the sequences $\{a_{n}\}$, $\{b_{n}\}$, and $\{c_{n}\}$ are convergent and $a_{n}\leq b_{n}\leq c_{n}$ is true always after some $n^{\mathrm {th} }$ term, if $\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }c_{n}=L$, then $\lim _{n\to \infty }b_{n}=L$.

Example 1

Determine whether the sequence $\left\{{\frac {x^{2}-1}{x+1}}\right\}_{n=1}^{\infty }$ is convergent or divergent.

Let $f(n)={\frac {n^{2}-1}{n+1}}$ be a function analogous to our sequence. When we factor the numerator and cancel like-terms, we get that $\lim _{n\to \infty }f(n)=\lim _{n\to \infty }{\frac {n^{2}-1}{n+1}}=\lim _{n\to \infty }{\frac {(n+1)(n-1)}{n+1}}=\lim _{n\to \infty }n-1=\infty$. Therefore the sequence $\left\{{\frac {x^{2}-1}{x+1}}\right\}_{n=1}^{\infty }$ is divergent.

Licensing

Content obtained and/or adapted from:

• Limit of a Sequence, mathonline.wikidot.com under a CC BY-SA license