Next we discover what sequences such as $\displaystyle \frac{1}{n^2}$ and $\displaystyle \frac{1}{n^3}$ tend to. This will be very useful later when combined with other techniques.

**Theorem**

Let $p\in \N$. Then $\displaystyle \lim_{n\to \infty} n^{-p}\to 0$.

**Proof**

Suppose $\varepsilon >0$ is given. Then, $n>\varepsilon ^{-1/p}$ implies that $n^{-p}<\varepsilon$. Thus let $N$ be an integer greater than $\varepsilon ^{-1/p}$. For all $n>N$ we have $|n^{-p}-0|=n^{-p}\lt \varepsilon $ so $\displaystyle \lim_{n\to \infty} n^{-p}\to 0$.

▢

**Example**

Consider the sequence $\displaystyle a_n=\frac{(-1)^n}{n^3}$. Then,

\[

\left| a_n \right|

=\left| \frac{(-1)^n}{n^3} \right|

=\frac{1}{n^3}\to 0.

\]

Thus, $\displaystyle a_n=\frac{(-1)^n}{n^3}\to 0$.

**Remark**

In the theorem we can change the assumption ‘$p\in \N$’ to ‘$p\in \R$ with $p>0$’.

For example, $\displaystyle \frac{1}{n^\pi }$ tends to $0$.

To see the difficulty consider what $2^\pi $ means? How do we multiply 2 by itself $\pi $ times? We know how to multiply $2$ by itself $3$ times and $4$ times but what about somewhere inbetween?

Thus, the general statement relies on defining $x^a$ where $a$ is a real number and $x>0$. You may have seen that one solution to this is to define $x^a$ (for $x>0$) to be $\exp(a\ln x)$ where $\exp $ and $\ln $ are the exponential and natural logarithm functions. These functions and $x^a$ are not defined rigorously yet. Hence, for the moment we shall restrict to the case of $p\in \N$.

If $a_n\to a$, then $|a_n|\to |a|$.

**Proof **

Given $\varepsilon >0$, there exists $N$ such that $|a_n-a|\lt \varepsilon $ for all $n>N$. But by the reverse triangle inequality

\[

\left| |a_n|-|a| \right| \leq |a_n-a| \lt \varepsilon {\text{ for all }} n>N.

\]

Hence, $|a_n|\to |a|$.

▢

This allows us to prove a simple result which turns out to be very useful.

**Corollary **

If $a_n\to a$ and $a_n\geq 0 $ for all $n\in \N$, then $a\geq 0$.

**Proof **

From the theorem we have that $a_n=|a_n|\to |a|$ and since limits are unique $a=|a|$, i.e., $a\geq 0$.

▢

We give an alternative proof for the corollary which rather than use the theorem uses ‘first principles’. I’ve included this as I have seen many first principles proofs in books and online that make it much more complicated than it needs to be.

**Alternative Proof **

Assume for a contradiction that $a\lt 0$. Then as $-a>0$, there exists $N$ such that

\[

|a_n-a|\lt -a {\text{ for all }} n>N.

\]

Then, as $a_n-a>0$ (because $a_n\geq0$ and $a\lt 0$), we have for $n>N$,

\[

a_n-a=|a_n-a|\lt -a ,

\]

i.e., $a_n\lt 0$, which is a contradiction.

Hence, $a\geq 0$.

▢

**How to think like a mathematician **

As usual we ask `What about the converse?’. Well, the converse to the theorem is not true. We can have $|a_n|\to |a|$ but $a_n\not\to a$. For example, $a_n=(-1)^n$.

The converse is true in a very special case (and we will make use of this a number of times):

**Theorem **

If $|a_n|\to 0$, then $a_n\to 0$.

**Proof **

Given $\varepsilon >0$, there exists an integer $N$ such that $n>N$ implies $\left| | a_n| – 0 \right|<\varepsilon $ for all $n>N$ since $|a_n|\to 0$.

But, $|a_n-0|=|a_n|=\left| | a_n| – 0 \right|\lt \varepsilon $ so $a_n\to 0$ also.

▢

What about sequences such as $1/2$, $1/4$, $1/8$, $1/16$, which can defined as $(1/2)^n$? The following proposition deals with this. It is very useful and is used countless times in mathematical proofs. The proof here uses properties of the log function, a function we have not defined rigorously but no doubt one you have met before.

**Proposition**

Suppose that $|x|\lt 1$. Then the sequence $x^n$ has limit $0$.

**Proof** If $x=0$, then we have a constant sequence and the limit is obviously $0$. Hence, we can assume $x\neq 0$.

Given $\varepsilon >0$, let $N\in \N$ be greater than $ \log \varepsilon / \log |x|$.

We have

\begin{eqnarray*}

|x^n-0|\lt \varepsilon &\iff &|x^n|<\varepsilon\\
&\iff & |x|^n \lt \varepsilon \\
&\iff & \log |x|^n \lt \log \varepsilon \\
&\iff & n\log |x| \lt \log \varepsilon \\
&\iff & n > \frac{\log \varepsilon }{\log |x| } .

\end{eqnarray*}

(The inequality sign reverses because $0\lt |x|\lt 1$ means $\log |x|$ is negative.)

Hence, if $n>N$, then $\displaystyle n> \frac{\log \varepsilon }{\log |x| } $ and so by the above we have $|x^n-0|\lt \varepsilon $. That is, $x^n$ has limit $0$.

▢

Note that in the definition of convergence that the definite article is used for the limit, i.e., we have ‘

First we need a lemma which should be intuitively obvious: if the difference between two numbers is less than all positive real numbers, then that difference is in fact zero and the numbers are equal.

**Lemma**

Suppose that $x$ and $y$ are real numbers such that $|x-y|\lt \varepsilon $ for all $\varepsilon >0$. Then $x=y$.

**Proof**

Suppose $x\neq y$. Then $|x-y|=\widetilde{\varepsilon}>0$ for some $\widetilde{\varepsilon}\in \R$. But then $|x-y|\lt \varepsilon $ is not true for $\varepsilon =\widetilde{\varepsilon}$ and this is a contradiction.

▢

Now for the observation that limits are unique.

**Theorem (Limits are unique)**

Suppose that $a_n \to a $ and $a_n \to a’ $. Then $a=a’$.

**Proof**

Let $\varepsilon >0$ be an arbitrary real number. Then there exists integers $N$ and $N’$ such that

\[

n>N \Longrightarrow |a_n-a|\lt \frac{\varepsilon }{2}

\]

and

\[

n>N’ \Longrightarrow |a_n-a’|\lt \frac{\varepsilon }{2} .

\]

(If the division by two is confusing, then see the remark following the proof.)

Hence, if $n>\max \{ N, N’ \} $, then

\begin{eqnarray*}

|a-a’|

&=& |a – a_n+ a_n -a’ | \\

&\leq & | a -a_n | + |a_n -a’| , {\text{ by the triangle inequality}}, \\

&=& |a_n -a| + |a_n -a’ | \\

&\lt & \frac{\varepsilon }{2} + \frac{\varepsilon }{2} \\

&=& \varepsilon .

\end{eqnarray*}

Since $\varepsilon $ was arbitrary, by the preceding lemma we deduce that $a=a’$.

▢

**Remark**

The use of $\varepsilon /2$ in the proof is confusing for many students. The objection is that the definition of limit says that given $\varepsilon >0$, there exists an $N$ such that $n>N\Longrightarrow |a_n-a|\lt \varepsilon $. Why can we divide the $\varepsilon $ by $2$?

Well, imagine that you have a machine that takes $\varepsilon$’s and produces $N$’s. Now suppose that someone gives you an $\varepsilon $. You can put $\varepsilon /2$ into the machine (instead of $\varepsilon$) and out will come an $N$ with $n>N\Longrightarrow |a_n-a|\lt \varepsilon /2$. In other words we have the situation that given $\varepsilon >0$, there exists $N$ such that $n>N\Longrightarrow |a_n-a|\lt \varepsilon /2$.

In fact, for a convergent sequence, given any $\varepsilon $ we can find $N$ such that $n>N\Longrightarrow |a_n-a|\lt \varepsilon /3$, or less than $\varepsilon /4$, or $7\varepsilon $, or $\sqrt{\varepsilon}$, or $\varepsilon ^{18}/25000$ or whatever.

Probably the most common mistake about the definition of limit is to describe it as `the thing that the sequence gets closer to but never reaches’. I have seen this so many times. The first mistake, of course, is that this is not the precise definition given. (Also, using the word ‘thing’ in any definition is a sure sign that there is something wrong!)

The second mistake is that it is not even true. Elements of the sequence can equal the limit and hence the limit is ‘reached’. Not only that, they can equal the limit in number of ways. In each of the following examples the limit is equal to zero.

- No elements equal the limit. E.g., $a_n=1/n$.
- A finite number of elements equal the limit. E.g., $a_n=\frac{1}{n}-\frac{3}{n^2}+\frac{2}{n^3}$.
- An infinite number of elements equal the limit. E.g., $a_n=\frac{\cos (n\pi /2 ) }{n}$.
- All elements equal the limit. E.g., $a_n=0$.

**Another Mistake**

A second, though less common, mistake is to think that the sequence steadily approaches a limit from one direction. True, we see many examples where this happens. For example, $1/n$ tends to its limit $0$ by approaching in a manner where each element is closer than the last.

This need not happen.

Consider the example above, $a_n=\frac{\cos (n\pi /2 ) }{n}$. Here the sequence ‘jumps about’. An element may be further from the limit than the one before it. However, as the sequence progresses the elements do get closer and closer.

**Notation mistakes**

A couple of other mistakes are essentially to do with notation.

- Error: ‘We have $a_n\to a$ for large $n$’.

This seems to be a confusion with approximations. E.g., $n^2+2n+3$ is approximately $n^2$ for large $n$.True, in the concept of limit we are investigating what happens as $n$ gets large but the correct way to say the above is ‘We have $a_n\to a$ as $n\to \infty $’.

- Error: ‘$\displaystyle \lim_{n\to\infty} a_n\to a$’. The error here is that the left hand side is a single
*number*not a sequence. We can only say a sequence tends to $a$. We can’t say that a number tends to $a$.The correct way is to write: $\displaystyle \lim_{n\to\infty} a_n=a$.

Let’s now see some examples of sequences that do not converge, i.e., they are

**Example**

The sequence $a_n=(-1)^n$ is not convergent.

To show that $a_n$ does not have a limit we shall assume, for a contradiction, that it does. Let $a$ be the limit of $(-1)^n$ as $n\to \infty $. Let $\varepsilon =1$. Then, there exists $N$ such that $|(-1)^n -a|\lt 1$ for all $n>N$. For an even $n$ we have $|1-a|\lt 1$ and for an odd $n$ we have $|1+a|=|-1-a|\lt 1$. Then, by the triangle inequality we have,

\[

2=|1-a+a+1|\leq | 1-a| +|1+a|\lt 1+1=2.

\]

This is a contradiction since $2\lt 2$ is definitely false. Hence, $(-1)^n$ has no limit as $n\to \infty $.

**Remark**

What is confusing for many students is that we chose $\varepsilon$ in the example. In our previous examples we had to deal with *all* $\varepsilon $ greater than zero — we had no choice in the matter. This is because we are assuming that the sequence is convergent and hence for every $\varepsilon >0$ we can find some $N\in \N$ so that $n>N\Longrightarrow |a_n-a|\lt \varepsilon$. Now, if we can find an $\varepsilon $ so that this last implication is false we get a contradiction — which is what we seek. We only need to find one such $\varepsilon $. It does not have to be all $\varepsilon >0$. (After all, to show a statement is false we only need one counterexample.)

**Example**

The sequence $a_n=n$ has no limit.

Suppose, for a contradiction, that $a_n$ has a limit $a$. Let $\varepsilon =1$. (Note that again we are choosing $\varepsilon $. That in our two examples we choose $1$ is in some sense a coincidence — it is a nice simple number.)

As $a_n$ converges there exists an $N$ so that $n>N\Longrightarrow |a_n-a|\lt 1$. But for $n>a+1$ we have $|a_n-a|=n-a>1$ which contradicts $|a_n-a|\lt 1$. Hence for any $n>\max\{N,a\}$ we have a contradiction.

The terms of the sequence $a_n=n$ get larger and larger. We shall introduce a notion in later to deal with sequences which ‘go to infinity’.

**Example**

Show that $a_n = \frac{3n^2}{n^2-5} $ has limit equal to $3$.

We do the usual and consider $|a_n-a|$:

\begin{align*}

|a_n-3|&= \left| \frac{3n^2}{n^2-5} – 3 \right| \\

&= \left| \frac{3n^2-3n^2+15}{n^2-5} \right| \\

&= \left| \frac{15}{n^2-5} \right| \\

&= \frac{15}{\left| n^2-5 \right|} .

\end{align*}

Now for $n^2-5>0$, i.e., $n>\sqrt{5}$ we have $\left| n^2-5 \right| >n^2-5$. Thus,

\[

|a_n-a| = \frac{15}{n^2-5} {\text{ for }} n>\sqrt{5} .

\]

Now, let’s introduce $\varepsilon$ (and keep assuming $ n>\sqrt{5}$):

\begin{align*}

\frac{15}{n^2-5}\lt \varepsilon

&\iff \frac{15}{\varepsilon} \lt n^2-5 \\

&\iff \sqrt{\frac{15}{\varepsilon} +5} \lt n .

\end{align*}

We have two conditions that we need $n$ to satisfy: We want $n>\sqrt{5}$ and $n>\sqrt{\frac{15}{\varepsilon} +5}$. Thus we take

\[

N>\max \left\{ \sqrt{5} , \sqrt{\frac{15}{\varepsilon} +5} \right\} = \sqrt{\frac{15}{\varepsilon} +5}.

\]

As usual let’s rewrite everything in a polished version.

**Polished version: **

Show that $a_n = \frac{3n^2}{n^2-5} \to 3$.

Given $\varepsilon >0$, let $N$ be an integer such that $N> \sqrt{\frac{15}{\varepsilon} +5}$. Then for $n>N$

\begin{align*}

|a_n-3|&= \left| \frac{3n^2}{n^2-5} – 3 \right|

= \left| \frac{3n^2-3n^2+15}{n^2-5} \right| \\

&= \left| \frac{15}{n^2-5} \right|

= \frac{15}{\left| n^2-5 \right|} \\

&=\frac{15}{n^2-5} , {\text{ as }} n>\sqrt{5} ,\\

&\lt \varepsilon ,{\text{ as }} n> \sqrt{\frac{15}{\varepsilon} +5} .

\end{align*}

Hence $a_n\to 3$.

Using the Algebra of Limits there is far superior way to prove that the limit is $3$.

**Remark**

It should be noted that if we have found a value for $N$, then any value bigger than that will also make the important condition hold.

So, for example, if $|a_n-a|\lt \varepsilon $ is true for all $n>5$, then clearly it is true for $n>6$, $n>7$, and $n>1,000,000$. In the above example, this means that we can remove the square root symbol. This is because

\[

N>\frac{15}{\varepsilon} +5 \Longrightarrow N>\sqrt{\frac{15}{\varepsilon} +5}

\]

We can go bigger still:

\[

N>\frac{15}{\varepsilon} +15 \Longrightarrow N>\frac{15}{\varepsilon} +5

\]

and so we could use any natural number $N$ greater than

$15 \left( 1+ \frac{1}{\varepsilon} \right) $.

By using the triangle inequality we can simplify expressions when we have negative signs in the expression for the sequence.

**Example**

Show that $a_n=\frac{2n^4+5n^3-3n^2+1}{n^4} \to 2$.

Again we simplify $|a_n-a|$ and find expressions that it is less than or equal to:

\begin{align*}

|a_n-2|&= \left| \frac{2n^4+5n^3-3n^2+1}{n^4} – 2 \right|

= \left| \frac{5n^3-3n^2+1}{n^4} \right| \\

&\leq \left| \frac{5n^3}{n^4} \right| + \left|\frac{-3n^2}{n^4} \right| + \left| \frac{1}{n^4} \right|, {\text{ by the triangle inequality}},\\

&= \frac{5n^3+3n^2+1}{n^4} \\

&\leq \frac{5n^3+3n^3+n^3}{n^4} , {\text{ for }} n\geq 1, ({\text{as }} n^2\leq n^3 {\text{ and }} 1\leq n^3),\\

&= \frac{9n^3}{n^4} \\

&=\frac{9}{n} .

\end{align*}

Hence we should take $N$ to be an integer greater than $9/\varepsilon $.

As usual let’s carefully write a coherent answer:

**Polished version: **

Show that $a_n=\frac{2n^4+5n^3-3n^2+1}{n^4} \to 2$.

Answer: Given $\varepsilon >0$, let $N$ be an integer greater than $9/\varepsilon$. Then for all $n>N$ we have $n>9/\varepsilon $, i.e, $\varepsilon >9/n$ and hence

\begin{align*}

|a_n-2|&= \left| \frac{2n^4+5n^3-3n^2+1}{n^4} – 2 \right|

= \left| \frac{5n^3-3n^2+1}{n^4} \right| \\

&\leq \left| \frac{5n^3}{n^4} \right| + \left|\frac{-3n^2}{n^4} \right| + \left| \frac{1}{n^4} \right|, {\text{ by the triangle inequality}},\\

&= \frac{5n^3+3n^2+1}{n^4} \\

&\leq \frac{5n^3+3n^3+n^3}{n^4} , {\text{ as }} n\geq 1,\\

&= \frac{9n^3}{n^4} \\

&=\frac{9}{n} \\

&\lt \varepsilon .

\end{align*}

Let’s try a more complicated example.

**Example**

Show that $a_n=\frac{2n+3}{n^2}$ has limit $0$.

We proceed as before by simplifying $|a_n-a|$ via equalities:

\[

|a_n-0|=\left| \frac{2n+3}{n^2} – 0 \right|

= \left| \frac{2n+3}{n^2} \right|

= \frac{2n+3}{n^2} .

\]

Now what we want is to find $N$ such that

\[

\frac{2n+3}{n^2} <\varepsilon {\text{ for all }} n>N.

\]

It is possible but messy to make $n$ the subject of this inequality so that we have $n$ greater than something. However, we shall adopt a simpler method by allowing ourselves to use $\leq $ and $<$ as follows:
\begin{align*}
\frac{2n+3}{n^2} &\leq \frac{2n+3n}{n^2}, {\text{ as obviously }} 3\leq 3n {\text{ for }} n\geq 1,\\
&=\frac{5n}{n^2}\\
&=\frac{5}{n} .
\end{align*}
This is much simpler. It doesn't have a quadratic term in $n$ and it is easy to find $N$ such that $n>N$ implies that $5/n\lt \varepsilon $, we just need $N>5/\varepsilon $.

Let’s put everything together in a coherent argument.

**Polished version: **

Show that $a_n=\frac{2n+3}{n^2}$ has limit $0$.

Answer: Given $\varepsilon >0$, let $N$ be an integer greater than $5/\varepsilon$. Then for all $n>N$ we have $n>5/\varepsilon $, i.e., $\varepsilon >5/n$ and hence

\[

|a_n-0|=\left| \frac{2n+3}{n^2} – 0 \right|

= \frac{2n+3}{n^2}

\leq \frac{2n+3n}{n^2}

=\frac{5n}{n^2}

=\frac{5}{n}

\lt \varepsilon .

\]

Notice that the written answer is short and the important choice, that of $N$, appears to come from nowhere. The reader does not need to know the working to find it. For the working we ensure that $|a_n-a|$ is less than, or less than or equal to, some simple expression.

Suppose that $a_n=c$ for all $n\in \N$. That is, $(a_n)$ is a constant sequence. Then, $\displaystyle \lim_{n\to \infty} a_n=c$.

**Proof**

Given any $\varepsilon$, let $N=1$. Then, for all $n>N$ we have

\[

|a_n-c|=|c-c|=0\lt \varepsilon .

\]