# Limits of Sequences

The **limit of a sequence** is the value the sequence approaches as the number of terms goes to infinity. Not every sequence has this behavior: those that do are called convergent, while those that don't are called divergent.

Limits capture the long-term behavior of a sequence and are thus very useful in bounding them. They also crop up frequently in real analysis.

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## Convergence of Sequences

Here, we will be discussing the aspects you will need to know for understanding the concept of convergence of a sequence. We will be delivering you a step-by-step presentation of all the concepts. First, what exactly is a sequence?

A **sequence** is a function $f : \mathbb N \rightarrow \mathbb R$ defined as $f(n) = x_n$, and is usually denoted by $x_1,x_2,..., x_n$ or simply by $x_n.$ We call $x_n$ the $n^\text{th}$ term of the sequence or the value of the sequence at $n.$ For example,

- $f(n)=\frac 1n$ generates the sequence $\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \ldots;$
- $f(n)=\frac{(-1)^n}{n}$ generates the sequence $\frac{-1}{1}, \frac 12, \frac{-1}{3} , \ldots;$
- $f(n)=1+\frac{1}{10^n}$ generates the sequence $1.1,1.01,1.001, \ldots;$
- $f(n)=(-1)^{n+1}$ generates the sequence $1, -1, 1, \ldots;$
- $f(n)=(1)^n+(-1)^n$ generates the sequence $0,2,0, \ldots;$
- $f(n)=n$ generates the sequence $1,2,3,\ldots;$
- $f(n)=\frac{n^3}{n+1}$ generates the sequence $\frac{1^3}{1+1}, \frac{2^3}{2+1}, \frac{3^3}{3+1}, \ldots$.

As we are now familiar with sequences, let us try to understand what the limit of a sequence represents. In simple words, a limit is a mathematically precise way to talk about approaching a value, without having to evaluate it directly.

A real number $L$ is the

limit of the sequence$x_n$ if the numbers in the sequence become closer and closer to $L$ and not to any other number. In a general sense, the limit of a sequence is the value that it approaches with arbitrary closeness.

For example, if $x_n = c$ for some constant $c,$ then $\displaystyle \lim_{n \to \infty} x_n \to c,$ and if $x_n = \frac 1n,$ then $\displaystyle \lim_{n \to \infty} x_n \to 0$.

When the limit of a sequence as $n \to \infty$ approaches a single value, we say the sequence converges. Let's define the convergence of a sequence in a formal way:

We say that a sequence $x_n$

convergesif there exists $x_0 \in \mathbb R$ such that for every $\epsilon> 0$, there exists a positive integer $N$ such that $x_n \in (x_0 - \epsilon,x_0 +\epsilon)$ or $|x_n -x_0| < \epsilon$ for all $n \geq N$.

It can be easily veriﬁed that if such a number $x_0$ exists then it is unique. In this case, we say that the sequence $x_n$ converges to $x_0$ and we call $x_0$ the limit of the sequence $x_n$. If $x_0$ is the limit of $x_n$, we write $\displaystyle \lim_{n \to \infty} x_n = x_0$.

Remark:The convergence of each sequence given in the above examples is veriﬁed directly from the deﬁnition. In general, verifying the convergence directly from the deﬁnition is a difficult task. We will see some methods to ﬁnd the limits of certain sequences and some sufficient conditions for the convergence of a sequence.

Now that we got the concept of convergence in theoretical terms, it's time now to work out some examples and build a strong foundation of the convergence of sequences. Here we go::

Does the following sequence converge:

$\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \ldots , \frac{1}{n}, \ldots \,?$

The sequence seems to be approaching 0. The larger $n$ gets, the smaller and smaller the term becomes towards 0. Thus, the sequence converges. $_\square$

Proof:

For arbitrary $\epsilon > 0$, the inequality $|x_n| = \frac 1n < \epsilon$ is true for all $n > \frac{1}{\epsilon}$ and thus for all $n > N$, where $N$ is any positive integer such that $N > \frac{1}{\epsilon}$. Thus for any $\epsilon > 0$, there is a positive integer $N$ such that $|x_n| <\epsilon$ for every $n \geq N$. $_\square$

Does the following sequence generated by the function $f(n)=1+\frac{1}{10^n}$ converge:

$1.1,\ 1.01,\ 1.001,\ \ldots \, ?$

In this sequence, we see that the values are decreasing as $n$ increases, and eventually approach a single value. The larger we take the value of $n$, the closer and closer the term becomes to 1. Hence the elements of the given sequence approach 1 when $n$ approaches infinity. So the sequence converges to 1. $_\square$

Does the following sequence generated by the function $f(n)=\frac{(-1)^n}{n}$ converge:

$\dfrac{-1}{1}, \dfrac 12, \dfrac{-1}{3} , \ldots \, ?$

The sequence seems to be approaching 0. The larger $n$ gets, the closer the term gets to 0. Thus, the sequence converges. Though the elements of the sequence $\frac{(-1)^n}{n}$ oscillate, they “eventually approach” the single point 0. The common feature of these sequences is that the terms of each sequence “accumulate” at only one point. $_\square$

Now, let's define the divergence of a sequence:

We say that a functiondivergesif the limit does not exist.

Does the following sequence converge:

$1, -1, 1, -1, 1, -1, \ldots, (-1)^{n+1}, \ldots \, ?$

It is clear that the sequence bounces back and forth between 1 and -1, and it doesn't converge down to a value. We say that the sequence diverges. The elements of the sequence $(-1)^n$ oscillate between two different points −1 and 1, which means the elements of the sequence come close to −1 and 1 “frequently” as $n$ increases. $_\square$

We say that a function

diverges to infinity, if it tends to positive infinity or negative infinity.

For an example, $f(n)=n$ and $f(n)=\ln n$ are such functions.

Does the following sequence converge:

$1, 2, 3, \ldots, n, \ldots \, ?$

The sequence of integers is unbounded above. Such sequences would diverge to (positive) infinity. The values of the sequence become larger and larger and do not accumulate anywhere. $_\square$

Does the following sequence converge:

$\dfrac{1^3}{1+1}, \dfrac{2^3}{2+1}, \dfrac{3^3}{3+1}, \ldots, \dfrac{n^3}{n+1} , \ldots \, ?$

**Note 1**: In the above examples, we see that if the difference between successive terms is bounded below by a constant $\big($2 in the $1^\text{st}$ example, 1 in the $2^\text{nd}$ example$\big),$ then such a sequence diverges. It will be shown below that if a sequence converges, then the limit of the difference between successive terms is 0.

**Note 2**: It is true that if a positive sequence is non-decreasing, then the limit exists. However, we might not be able to easily determine the limit.

## Graphical Examples

Graphical interpretation of sequences is an easy tool to determine convergence:

- sometimes it is easy to see;
- sometimes we might draw the wrong conclusion, e.g. $\ln n.$

Find the limit

$\lim_{n \to \infty} \cos n\pi.$

Let's evaluate the first few terms of this sequence.

- For $n = 1,$ $\cos n \pi = \cos 1\pi = -1$.
- For $n = 2,$ $\cos n \pi = \cos 2\pi = 1$.
- For $n = 3,$ $\cos n \pi = \cos 3\pi = -1$.
- For $n = 4,$ $\cos n \pi = \cos 4\pi = 1$.
- For $n = 5,$ $\cos n \pi = \cos 5\pi = -1$.
Since the terms of the sequence oscillate between -1 and 1, we can conclude the sequence diverges or doesn't converge down to a single value. $_\square$

Find the limit of the sequence

$\frac{1}{ \sqrt{1} } , \frac{1}{ \sqrt{2} }, \frac{ 1 } { \sqrt{3} }, \ldots , \frac{1}{ \sqrt{n} }, \ldots .$

If we write out the initial few terms, we will get $1, 0.707\ldots, 0.577 \ldots,$ $0.5, 0.445\ldots, 0.408, \ldots,$ and so on. It is not immediately apparent what the limit is.

Let's think about what happens when $n$ is really large.

- If $n > 100$, then $\sqrt{n} > 10,$ so $\frac{1} { \sqrt{n} } < \frac{1}{10} =0.1$.
- If $n > 10000$, then $\sqrt{n} > 100,$ so $\frac{1} { \sqrt{n} } < \frac{1}{100} =0.01$.
- If $n > 1000000$, then $\sqrt{n} > 1000,$ so $\frac{1} { \sqrt{n} } < \frac{1}{1000} =0.001$.
Thus, the limit of the sequence is 0. $_\square$

## Using Properties of Limits

You should be familiar with the following properties of limits. If the limits $\lim a_n$ and $\lim b_n$ exist and are finite, then

$\begin{array} { l r l } 1. & \lim_{n \to \infty}\left( a_n \pm b_n \right) &= \lim_{n \to \infty} a_n \pm \lim_{n \to \infty} b_n \\ 2. & \lim_{n \to \infty} (c\cdot a_n) &= c \cdot \lim_{n \to \infty} a_n \\ 3. & \lim_{n \to \infty} \left( a_n b_n \right) &= \Big( \lim_{n \to \infty} a_n \Big) \Big(\lim_{n \to \infty} b_n \Big) \\ 4. & \lim_{n \to \infty} \frac{a_n}{b_n} &=\frac{\displaystyle \lim_{n \to \infty} a_n}{\displaystyle \lim_{n \to \infty} b_n} \text{, as long as } \lim_{n \to \infty} b_n \neq 0. \end{array}$

What is the limit of the sequence

$\frac{1}{2} , \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \ldots , \frac{n}{n+1}, \ldots \, ?$

To start, let's list out the terms.

- For $n = 1$, $\frac{1}{1+1} = \frac{1}{2} = 0.5$.
- For $n = 2$, $\frac{2}{2+1} = \frac{2}{3} = 0.666\ldots$.
- For $n = 3$, $\frac{3}{3+1} = \frac{3}{4} = 0.75$.
- For $n = 4$, $\frac{4}{4+1} = \frac{4}{5} = 0.8$.
- For $n = 5$, $\frac{5}{5+1} = \frac{5}{6} = 0.866\ldots$.
We see that the terms are increasing, and seem to be getting close to 1.

Notice that another way of writing the sequence is as $1 - \frac{1}{n+1}$. We know that the limit of the constant 1 is just 1, and the limit of $\frac{1}{n+1}$ is 0, so we can apply the first rule to conclude that

$\lim_{n \rightarrow \infty } \frac{n}{n+1} = \lim_{n \rightarrow \infty } 1 - \frac{1}{n+1} = \lim_{n \rightarrow \infty } 1 - \lim_{n \rightarrow \infty } \frac{1}{n+1} = 1 - 0 = 1 . \ _\square$

Find $\displaystyle \lim_{n \to \infty} \frac{1}{n^2}.$

We know that $\displaystyle \lim_{n \to \infty} \frac{1}{n}=0.$ Therefore, by applying the third rule, we have

$\begin{aligned} \lim_{n \to \infty} \frac{1}{n^2} &= \lim_{n \to \infty} \left(\frac{1}{n} \times \frac{1}{n}\right) \\ &= \left(\lim_{n \to \infty} \frac{1}{n}\right)\left(\lim_{n \to \infty} \frac{1}{n}\right) \\ &= 0 \times 0 \\ &= 0. \ _\square \end{aligned}$

Find $\displaystyle \lim_{n \to \infty} \frac{n^2+5n}{3n^2+1}.$

By factoring the term of highest degree from both the numerator and denominator, we have $\lim_{n \to \infty} \frac{n^2+5n}{3n^2+1} = \lim_{n \to \infty} \frac{1+\frac{5}{n}}{3+\frac{1}{n^2}}.$ Now, by applying the results of previous examples that $\displaystyle \lim_{n \to \infty} \frac{1}{n}=0$ and $\displaystyle \lim_{n \to \infty} \frac{1}{n^2}=0,$ we have $\begin{aligned} \lim_{n \to \infty} \left(1+\frac{5}{n}\right) &= \lim_{n \to \infty} (1)+5\lim_{n \to \infty} \frac{1}{n}=1 \text{ and} \\ \lim_{n \to \infty} \left(3+\frac{1}{n^2}\right) &= \lim_{n \to \infty} (3)+\lim_{n \to \infty} \frac{1}{n^2}=3. \end{aligned}$ Therefore, $\lim_{n \to \infty} \frac{n^2+5n}{3n^2+1}=\frac{\displaystyle \lim_{n \to \infty} \left(1+\frac{5}{n}\right)}{\displaystyle \lim_{n \to \infty} \left(3+\frac{1}{n^2}\right)}=\frac{1}{3}. \ _\square$

Find $\displaystyle \lim_{n \to \infty} \left(\log_{10} \big(10n^2-2n\big)-\log_{10} \big(n^2+1\big)\right).$

By using the property of logarithm that $\log_a x-\log_a y=log_a \frac{x}{y},$ we can rewrite the given equation and get the limit value as follows:

$\begin{aligned} \lim_{n \to \infty} \left(\log_{10} \big(10n^2-2n\big)-\log_{10} \big(n^2+1\big)\right) &= \lim_{n \to \infty} \log_{10} \frac{10n^2-2n}{n^2+1} \\ &= \lim_{n \to \infty} \log_{10} \frac{10-\frac{2}{n}}{1+\frac{1}{n^2}} \\ &= \log_{10} \frac{10-0}{1+0} \\ &= \log_{10} 10 \\ &= 1. \ _\square \end{aligned}$

## Find $\displaystyle \lim_{n \to \infty} \frac{1^2+2^2+3^2+\cdots +n^2}{n^3+2}.$

Observe that $\begin{aligned} 1^2+2^2+3^2+\cdots +n^2 &= \sum_{k=1}^{n} k^2 \\ &= \frac{n(n+1)(2n+1)}{6}, \end{aligned}$ then the value of the given equation can be calculated as follows: $\begin{aligned} \lim_{n \to \infty} \frac{1^2+2^2+3^2+\cdots +n^2}{n^3+2} &= \lim_{n \to \infty} \frac{1}{n^3+2} \cdot \frac{n(n+1)(2n+1)}{6} \\ &= \lim_{n \to \infty} \frac{2n^3+3n^2+n}{6n^3+12} \\ &= \lim_{n \to \infty} \frac{2+\frac{3}{n}+\frac{1}{n^2}}{6+\frac{12}{n^3}} \\ &= \frac{2}{6} \\ &= \frac{1}{3}. \ _\square \end{aligned}$

For a positive integer $n,$ let $a_n$ be the fractional part of $\sqrt{n^2+5n+4}.$ Then find $\displaystyle \lim_{n \to \infty} a_n.$

For a positive integer, it must be true that $n^2+4n+4 < n^2+5n+4 < n^2+6n+9,$ which implies $(n+2)^2 < n^2+5n+4 < (n+3)^2.$ Hence, we have $n+2 < \sqrt{n^2+5n+4} < n+3.$ Therefore, the integer part of $\sqrt{n^2+5n+4}$ is $(n+2)$ and the fractional part of it is $a_n=\sqrt{n^2+5n+4}-(n+2).$ Thus, we have $\begin{aligned} \lim_{n \to \infty} a_n &= \lim_{n \to \infty} \left( \sqrt{n^2+5n+4}-(n+2) \right) \\ &= \lim_{n \to \infty} \frac{\left(\sqrt{n^2+5n+4}-(n+2)\right)\left(\sqrt{n^2+5n+4}+(n+2)\right)}{\sqrt{n^2+5n+4}+(n+2)} \\ &= \lim_{n \to \infty} \frac{n}{\sqrt{n^2+5n+4}+(n+2)} \\ &= \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{5}{n}+\frac{4}{n^2}}+1+\frac{2}{n}} \\ &= \frac{1}{\sqrt{1+0+0}+1+0} \\ &= \frac{1}{2}. \ _\square \end{aligned}$

## Epsilon-delta Definition

Main Article: Epsilon-Delta Definition of a Limit

Stated precisely, the epsilon-delta definition of a limit is $\displaystyle \lim_{n \to \infty } \left\{ a_n \right\}=L$ if for every $\epsilon > 0$ there exists a positive integer $M$ such that

$\text{if } n > M \text{, then } \left| a_n - L \right| < \epsilon .$

If the sequence $a_n$ converges, then

$\lim_{n\rightarrow \infty} a_{n+1} - a_n = 0.$

**Note:** The converse of the theorem is not true. For example, consider the sequence $\displaystyle a_n = \sum_{i=1} ^ n \frac{1}{i}$, which is the sum of the reciprocals. The difference of successive terms is $\frac{1}{ i + 1 },$ which tends to 0. However, the sum of reciprocals diverges to infinity.

**Cite as:**Limits of Sequences.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/limits-of-sequences/