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.

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,

1. $f(n)=\frac 1n$ generates the sequence $\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \ldots;$
2. $f(n)=\frac{(-1)^n}{n}$ generates the sequence $\frac{-1}{1}, \frac 12, \frac{-1}{3} , \ldots;$
3. $f(n)=1+\frac{1}{10^n}$ generates the sequence $1.1,1.01,1.001, \ldots;$
4. $f(n)=(-1)^{n+1}$ generates the sequence $1, -1, 1, \ldots;$
5. $f(n)=(1)^n+(-1)^n$ generates the sequence $0,2,0, \ldots;$
6. $f(n)=n$ generates the sequence $1,2,3,\ldots;$
7. $f(n)=\frac{n^3}{n+1}$ 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 $x_n$ converges if 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 function diverges if 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 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. $_\square$

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 $$\begin{aligned} \lim_{n \to \infty} a_n &= \lim_{n \to \infty} \left( \sqrt{n^2+5n+4}-(n+2) \right) \\ &= \lim_{n \to \infty} \frac{\left(\sqrt{n^2+5n+4}-(n+2)\right)\left(\sqrt{n^2+5n+4}+(n+2)\right)}{\sqrt{n^2+5n+4}+(n+2)} \\ &= \lim_{n \to \infty} \frac{n}{\sqrt{n^2+5n+4}+(n+2)} \\ &= \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{5}{n}+\frac{4}{n^2}}+1+\frac{2}{n}} \\ &= \frac{1}{\sqrt{1+0+0}+1+0} \\ &= \frac{1}{2}. \ _\square \end{aligned}$$

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.