Limits of Sequences
Introduction
Mathematical analysis studies the infinitesimal properties of functions. It is the entire theory whose formal axiomatic introduction requires a lot of time. If you are reading these pages, you are probably not looking for that. You want to gain or improve your skills in solving problems in calculus.
The aim of this set of notes is to introduce you to calculus without rigorously building the theory of mathematical analysis. Some of the proofs will be omitted or their details will be skipped. However, we will rigorously define the concept of limits of sequences. This way we will provide convincing arguments that the theory of limits and integration can be developed in the same way as number theory, combinatorics, and geometry. You will also get an insight of how proofs work in mathematical analysis. However, the main goal is to provide you with techniques for differentiation and integration and after this introductory chapter you may notice a more serious lack of details and rigor. This chapter is an exception. It is more mathematical analysis than it is calculus, but even if you think that that’s not what you want, please take your time to read this. And try to enjoy.
Motivation
We will define the limit of a sequence and establish some important properties of limits.
For motivational purposes we are going to give the following fake story of the history of number \(\pi\):
Million years ago people decided that it is important to find out the ratio between the perimeter and the diameter of a circle. The first calculations were not precise because people could make relatively small circles. They got that the ratio is \(3\). Let us denote by \(\rho_1\) this first primitive ratio that the society has obtained. So, we got our first equation: \(\rho_1=3\).
After some time, they built a better technology  i.e. the rope. The rope made it possible to make bigger circles and get a more precise ratio of the perimeter to the radius. And they got \(\rho_2=3.1\).
After several years the price of rope decreased. Mathematicians got a big grant to buy a rope and they made their next scientific breakthrough: \(\rho_3=3.14\). Year after year they improved this ratio. They obtained the sequence
\[\rho_1=3,\;\;\; \rho_2=3.1, \;\;\; \rho_3=3.14,\;\;\; \rho_4=3.142,\;\;\; \rho_5=3.1416,\;\;\; \rho_6=3.14159,\;\;\; \rho_7=3.141593,\;\;\; \dots\]
They figured out at that point that no society will ever live to see the final product: the exact ratio written as a number with a finite decimal expansion. The sequence \((\rho_n)_{n=1}^{\infty}\) is going to be truly infinite sequence of numbers that are getting closer and closer to the real ratio. They gave the name \(\pi\) to the real ratio and said ``The sequence \(\left(\rho_n\right)_{n=1}^{\infty}\) converges to \(\pi\).’’
We want to define the notion of ``convergence.’’
Let us see a few more examples of sequences and further develop our intuition on convergence. The most trivial case is a constant sequence \((a_n)_{n=1}^{\infty}\) defined as \(a_n=0\). This sequence converges to \(0\). A typical divergent sequence occurs when there are two subsequences that approach to different limits. An example is the sequence \((b_n)_{n=1}^{\infty}\) defined as \[b_n=\left\{\begin{array}{ll} 0,&\mbox{ if }n \mbox{ is odd,}\newline
1,&\mbox{ if }n\mbox{ is even.}\end{array}\right.\]
Another example of divergent sequence is \((d_n)_{n=1}^{\infty}\) defined as:
\[d_n=\left\{\begin{array}{ll} \frac1n,&\mbox{ if }n \mbox{ is odd,}\newline
1+\frac1n,&\mbox{ if }n\mbox{ is even.}\end{array}\right.\]
Notice that the sequences \(\left(\frac1n\right)_{n=1}^{\infty}\) and \(\left(1+\frac1n\right)_{n=1}^{\infty}\) are both convergent.
Definition of the limit
The sequence whose elements are \(a_1\), \(a_2\), \(\dots\) will be referred to as \((a_n)_{n=1}^{\infty}\).
Definition A sequence \((a_n)_{n=1}^{\infty}\) of real numbers is called convergent if there exists a real number \(L\) that satisfies:
For each \(\varepsilon > 0\) there exists a positive integer \(n_0\in\mathbb N\) such that \(n\geq n_0\) implies \(\lefta_nL\right < \varepsilon\).
The real number \(L\) is called the limit of the sequence \((a_n)_{n=1}^{\infty}\) and we write this as \[\lim_{n\to\infty}a_n=L.\]
Example
Assume that \(a_n=\left(3+\frac1n\right)^2\). Prove that this sequence is convergent and find its limit.
We will prove that the limit of this sequence is \(9\). Assume that \(\varepsilon > 0\) is given. Then we can take \(n_0\) to be any integer greater than \(\frac{7}{\varepsilon}\). For such an \(n_0\) we have that \(n\geq n_0\) implies that \[\left\left(3+\frac1n\right)^29\right=\left\frac 6n+\frac1{n^2}\right=\frac6n+\frac1{n^2}\leq \frac6n+\frac1n < \varepsilon.\]
Main properties of limits
Theorem 1 (Convergent sequences are bounded)
Assume that \((a_n)_{n=1}^{\infty}\) is convergent. Then there exists a real number \(M\) such that \(a_n < M\) for all \(n\in\mathbb N\).
Let us set \(\varepsilon=1\). From the definition of the convergent sequence we see that there exists a real number \(L\) and an integer \(n_0\in\mathbb N\) such that \(a_nL < 1\) for \(n\geq n_0\). Let \(R=\max\left\{a_1L,a_2L,\dots, a_{n_0}L\right\}\). The number \(R\) is a real number (as opposed to \(+\infty\)) because it is a maximum of a finite set of numbers. Therefore, \(a_nL\leq \max\{1,R\}\) for all \(n\in\mathbb N\) hence: \[a_n=a_nL+L\leq a_nL+L\leq \max\{1,R\}+L,\] hence we may take \(M= \max\{1,R\}+L\).
Theorem 2 (Squeeze Theorem)
Assume that \((a_n)_{n=1}^{\infty}\), \((b_n)_{n=1}^{\infty}\), and \((c_n)_{n=1}^{\infty}\) are three sequences such that there exists \(n_0\in\mathbb N\) that satisfies
\[n\geq n_0\quad\quad\quad \mbox{implies}\quad\quad\quad a_n\leq b_n\leq c_n.\] If \(\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty} c_n=L\), then the sequence \((b_n)_{n=1}^{\infty}\) is convergent and \(\displaystyle\lim_{n\to\infty}b_n=L\).
Assume that \(\varepsilon > 0\) is given.
Since \(\lim_{n\to\infty}a_n=L\) there exists a positive integer \(n_1\) for which \(n\geq n_1\) implies that \(a_nL < \varepsilon\). Similarly, there exists \(n_2\) for which \(n\geq n_2\) implies that \(c_nL < \varepsilon\). Then if we define \(n_3=\max\{n_0,n_1,n_2\}\) we have that \(n\geq n_3\) implies that \(\varepsilon < a_nL\); \(a_n < b_n < c_n\), and \(c_nL < \varepsilon\). This means that \[L\varepsilon < a_n < b_n < c_n < L+\varepsilon\quad\quad\quad\Rightarrow \quad\quad\quad b_nL < \varepsilon.\]
The last inequality holds for every \(n\geq n_3\) hence \(\displaystyle\lim_{n\to\infty} b_n=L\).
Theorem 3 (Limit of the sum)
Assume that \((a_n)_{n=1}^{\infty}\) and \((b_n)_{n=1}^{\infty}\) are two convergent sequences whose limits are \(\alpha\) and \(\beta\), respectively. Then the sequence \((a_n+b_n)_{n=1}^{\infty}\) is convergent and its limit is \(\alpha+\beta\).
Assume that \(\varepsilon > 0\) is given. Since \(\displaystyle \lim_{n\to\infty}a_n=\alpha\) there exists \(n_1\in\mathbb N\) such that if \(n\geq n_1\) the following inequality holds: \(a_n\alpha < \frac{\varepsilon}2\). Similarly, there exists \(n_2\in\mathbb N\) such that \(n\geq n_2\) implies \(b_n\beta\leq \frac{\varepsilon}2\). Let us denote \(n_0=\max\{n_1,n_2\}\). If \(n\geq n_0\) then \(a_n\alpha < \frac{\varepsilon}2\) and \(b_n\beta\leq \frac{\varepsilon}2\) hence:
\[\lefta_n+b_n(\alpha+\beta)\right\leq \lefta_n\alpha\right+\leftb_n\beta\right\leq \frac{\varepsilon}2+\frac{\varepsilon}2=\varepsilon.\] The proof is complete.
Theorem 4 (Limit of the product)
Assume that \((a_n)_{n=1}^{\infty}\) and \((b_n)_{n=1}^{\infty}\) are two convergent sequences whose limits are \(\alpha\) and \(\beta\), respectively. Then the sequence \((a_n\cdot b_n)_{n=1}^{\infty}\) is convergent and its limit is \(\alpha\cdot \beta\).
Assume that \(\varepsilon > 0\) is given. Since \((a_n)_{n=1}^{\infty}\) and \((b_n)_{n=1}^{\infty}\) are convergent, these two sequences are bounded and there exist numbers \(M_a\) and \(M_b\) such that \(a_n\leq M_a\) and \(b_n\leq M_b\) for all \(n\in\mathbb N\). Let \(M=\max\{M_a,M_b\}+1\).
Since \(\displaystyle \lim_{n\to\infty}a_n=\alpha\) there exists \(n_1\in\mathbb N\) such that if \(n\geq n_1\) the following inequality holds: \(a_n\alpha < \frac{\varepsilon}{2M}\). Similarly, there exists \(n_2\in\mathbb N\) such that \(n\geq n_2\) implies \(b_n\beta\leq \frac{\varepsilon}{2M}\). Let us denote \(n_0=\max\{n_1,n_2\}\). If \(n\geq n_0\) then
\begin{eqnarray*}
a_n\cdot b_n\alpha\cdot \beta&=&a_nb_n\alpha b_n+\alpha b_n\alpha\beta\\&\leq& a_nb_n\alpha b_n+\alpha b_n\alpha\beta= b_n\cdot a_n\alpha+\alpha\cdot b_n\beta
\\ &\leq& M\cdot \frac{\varepsilon}{2M}+M\cdot \frac{\varepsilon}{2M}=\varepsilon.
\end{eqnarray*}
Thus \(\displaystyle \lim_{n\to\infty} a_nb_n=\alpha\beta\).
Theorem 5 (Limit of the quotient)
Assume that \((a_n)_{n=1}^{\infty}\) and \((b_n)_{n=1}^{\infty}\) are two convergent sequences whose limits are \(\alpha\) and \(\beta\), respectively. Assume that \(b_n\neq 0\) for all \(n\) and that \(\beta\neq 0\). Then the sequence \(\left(\frac{a_n}{b_n}\right)_{n=1}^{\infty}\) is convergent and its limit is \(\frac{\alpha}{\beta}\).
Assume that \(\varepsilon > 0\) is given. Since \(\displaystyle \lim_{n\to\infty}b_n=\beta\) and \(\beta\neq 0\), we may assume that \(\beta > 0\). Let us fix \(\beta^{\prime}\in(0,\beta)\). There exists \(n_1\in\mathbb N\) such that \(n\geq n_1\) implies that \(b_n > \beta^{\prime}\).
Since \((a_n)_{n=1}^{\infty}\) is convergent there exist a number \(M_a\) such that \(a_n\leq M_a\). Let \(M=\max\left\{\frac{M_a}{\beta^{\prime}\,^2},\frac1{\beta^{\prime}}\right\}+1\).
Since \(\displaystyle \lim_{n\to\infty}a_n=\alpha\) there exists \(n_2\in\mathbb N\) such that if \(n\geq n_2\) the following inequality holds: \(a_n\alpha < \frac{\varepsilon}{2M}\). Similarly, there exists \(n_3\in\mathbb N\) such that \(n\geq n_3\) implies \(b_n\beta\leq \frac{\varepsilon}{2M}\). Let us denote \(n_0=\max\{n_1,n_2,n_3\}\). If \(n\geq n_0\) then
\begin{eqnarray*}
\left\frac{a_n}{b_n}\frac{\alpha}{\beta}\right&=&\left\frac{a_n}{b_n}\frac{a_n}{\beta}+\frac{a_n}{\beta}\frac{\alpha}{\beta}\right\\
&\leq&
\left\frac{a_n}{b_n}\frac{a_n}{\beta}\right+\left\frac{a_n}{\beta}\frac{\alpha}{\beta}\right=
\left\frac{a_n}{b_n\cdot \beta}\right\cdotb_n\beta+\frac1{\beta}\cdot a_n\alpha\\
&< &M\cdot \frac{\varepsilon}{2M}+M\cdot \frac{\varepsilon}{2M}=\varepsilon.
\end{eqnarray*}
Thus \(\displaystyle \lim_{n\to\infty} \frac{a_n}{b_n}=\frac{\alpha}{\beta}\).
A sequence \((b_n)_{n=1}^{\infty}\) is called a subsequence of \((a_n)_{n=1}^{\infty}\) if there exists an increasing sequence of positive integers \(i_1\), \(i_2\), \(\dots\), such that \(b_n=a_{i_n}\).
Theorem 6 If \((a_n)_{n=1}^{\infty}\) is a convergent sequence with limit \(L\), then every subsequence of \((a_n)\) is convergent and its limit is \(L\).
Assume that \((b_n)_{n=1}^{\infty}\) is a subsequence of \((a_n)_{n=1}^{\infty}\). Assume that \(i_1\), \(i_2\), \(\dots\) is an increasing sequence of positive integers such that \(b_n=a_{i_n}\).
Given \(\varepsilon > 0\), let us choose \(n_0\) such that \(n\geq n_0\) implies \(a_nL < \varepsilon\). Then for \(n\geq n_0\) we are sure \(i_n\geq n\geq n_0\) since \((i_n)_{n=1}^{\infty}\) is increasing. Therefore \(b_nL=a_{i_n}L < \varepsilon\) hence
\(\displaystyle \lim_{n\to\infty}b_n=L\).
Theorem 7 If \((a_n)_{n=1}^{\infty}\) is a convergent sequence of positive real numbers whose limit is \(L\), then the sequence \((b_n)_{n=1}^{\infty}\) defined as \(b_n=\sqrt{a_n}\) is convergent as well and its limit is \(\sqrt L\).

Assume first that \(L > 0\).
Assume that \(\varepsilon > 0\) is given. There exists \(n_0\in\mathbb N\) such that \(n > n_0\) implies that \(a_nL < \varepsilon \cdot \sqrt L\).
We now have that for \(n\geq n_0\) the following relations hold:
\begin{eqnarray*}
b_n\sqrt L&=&\sqrt{a_n}\sqrt L=\frac{a_nL}{\sqrt{a_n}+\sqrt L}\leq\frac{a_nL}{\sqrt L}< \varepsilon.
\end{eqnarray*}

Assume now that \(L=0\).
There exists \(n_0\in\mathbb N\) such that \(n\geq n_0\) implies \(a_n < \varepsilon^2\).
This in turn implies that \(b_n=\sqrt{a_n} < \varepsilon\) which means that \(\displaystyle \lim_{n\to\infty} b_n=0=\sqrt L\).
The previous theorem can be written as \[\lim_{n\to\infty} \sqrt{a_n}=\sqrt{\lim_{n\to\infty}a_n}.\] provided that \((a_n)\) has nonnegative terms and that it is convergent. We will later see that for a class of functions (called continuous functions) we have \(\displaystyle \lim_{n\to\infty} f(a_n)=f\left(\lim_{n\to\infty}a_n\right)\).
The following theorems allow us to establish the existence of limits even in situations when we do not see what the limit of the sequence is. They will imply the existence but they won’t tell us what is the limit. These theorems played a central role in the development of mathematical analysis. The ideas behind their proofs are far reaching. They lead to the theory of Banach spaces.
Theorem 8 If the sequence \((a_n)_{n=1}^{\infty}\) is monotone and bounded, then it is convergent.
Assume that \(a_n\) is nondecreasing. That means \(a_{n+1}\geq a_n\). The case of nonincreasing sequence is the same.
Assume that \(M\) is a real number such that \(a_n\leq M\) for all \(n\in\mathbb N\). Consider the set \(A=\{a_1,a_2, a_3,\dots\}\). This set is bounded from above.
Among all upper bounds of \(A\) there exists the smallest one  let’s call it \(L\). Don’t ask for proof of this  this is an axiom. This smallest upper bound \(L\) is called the supremum of \(A\). Existence of the supremum for any bounded set is an axiom for real numbers!
We will now prove that \(L\) is the limit of the sequence \((a_n)_{n=1}^{\infty}\). Assume that \(\varepsilon > 0\) is given. The number \(L\varepsilon\) is smaller than the smallest upper bound. Therefore it can’t be an upper bound, and there exists an \(n_0\) such that \(a_{n_0} > L\varepsilon\). Since \((a_n)_{n=1}^{\infty}\) is nondecreasing we have that for every \(n\geq n_0\) the following inequality holds: \(L\varepsilon < a_{n_0}\leq a_n\). Since \(L\) is an upper bound we also have \(a_n < L\) therefore \(a_nL < \varepsilon\).
Theorem 9
For each bounded sequence \((a_n)_{n=1}^{\infty}\) we define the sequences \((\overline{a}_n)_{n=1}^{\infty}\) and \((\underline{a}_n)_{n=1}^{\infty}\) in the following way:
\begin{eqnarray*}
\overline{a}_n&=& \sup\left\{a_n,a_{n+1},\dots \right\},\\
\underline{a}_n&=&\inf\left\{a_n,a_{n+1},\dots\right\}.
\end{eqnarray*}

(a)
The sequences \((\overline{a}_n)_{n=1}^{\infty}\) and \((\underline{a}_n)_{n=1}^{\infty}\) are convergent and their limits are called \(\displaystyle \limsup_{n\to\infty} a_n\) and \(\displaystyle \liminf_{n\to\infty} a_n\).

(b) There are subsequences \((b_n)_{n=1}^{\infty}\) and \((c_n)_{n=1}^{\infty}\) of
\((a_n)_{n=1}^{\infty}\) such that \[\lim_{n\to\infty} b_n=\limsup_{n\to\infty} a_n\quad\quad\quad \mbox{and} \quad\quad\quad \lim_{n\to\infty} c_n=\liminf_{n\to\infty} a_n.\]
(c)
The sequence \((a_n)_{n=1}^{\infty}\) is convergent if and only if \[\limsup_{n\to\infty}a_n=\liminf_{n\to\infty} a_n.\]

(a)
The statement follows immediately from the previous theorem and the fact that \((\overline{a}_n)_{n=1}^{\infty}\) is decreasing and \((\underline{a}_n)_{n=1}^{\infty}\) is increasing.

(b) Let us denote \(\overline{a}=\limsup_{n\to\infty}a_n\) and \(\underline{a}=\liminf_{n\to\infty}a_n\).
Let us prove the existence of the sequence \((b_n)_{n=1}^{\infty}\). The existence of \((c_n)_{n=1}^{\infty}\) is established in an analogous way.
For each \(n\in\mathbb N\) there exists an element \(a_{k_n}\) of the set \(\{a_n, a_{n+1},\dots\}\) such that \(a_{k_n}\geq \overline{a}_n\frac1{n}\). We will set \(b_n=a_{k_n}\). We will now prove that \(\displaystyle \lim_{n\to\infty} b_n=\overline{a}\).
Let \(\varepsilon > 0\). There exists \(n_1\in\mathbb N\) such that \(n\geq n_1\) implies that \(\overline{a}_n\overline a < \frac{\varepsilon}2\). Let us define \(n_0=\max\left\{n_1,\left\lceil\frac2{\varepsilon}\right\rceil\right\}\). For \(n\geq n_0\) we have that \[\overline ab_n\leq \overline a\overline{a}_n+\overline{a}_nb_n\leq\frac{\varepsilon}2+\frac1n\leq \varepsilon.\]

(c) This part follows from Theorem 6 and part (b).
Theorem 10 (Cauchy’s Theorem) Assume that the sequence \((a_n)_{n=1}^{\infty}\) has the following property: For every \(\varepsilon > 0\) there exists \(n_0\) such that
the inequality \(a_na_m < \varepsilon\) holds for any two natural numbers \(m\) and \(n\) that satisfy \(m\geq n_0\) and \(n\geq n_0\).
Then the sequence \((a_n)_{n=1}^{\infty}\) is convergent.
We prove this theorem in three steps:
The sequenze \((a_n)_{n=1}^{\infty}\) is bounded:
Take \(\varepsilon=1\). There exists an integer \(n_0\) such that \(m,n\geq n_0\) implies that \(a_na_m < 1\). Let \[M=\max\left\{a_1,\dots, a_{n_0}\right\}.\] The last maximum is finite because we are taking the maximum of finitely many real number. Therefore for all \(n\geq n_0\) we have \[a_n=\lefta_na_{n_0}+a_{n_0}\right\leq \lefta_na_{n_0}\right+\lefta_{n_0}\right\leq 1+\lefta_{n_0}\right\leq 1+M.\]
The sequence \((a_n)_{n=1}^{\infty}\) has a convergent subsequence:
This follows immediately from the part (b) of Theorem 9.
Consider a convergent subsequence \((b_n)_{n=1}^{\infty}\) of \((a_n)_{n=1}^{\infty}\). Let \(L\) be the limit of \((b_n)_{n=1}^{\infty}\). We will prove that \(\lim_{n\to \infty} a_n=L\). Let \(\varepsilon > 0\).
There exists an integer \(n_1\) such that \(a_na_m < \frac{\varepsilon}2\) for \(n,m\geq n_1\). Let \(n^{\prime}\) be an integer bigger than \(n_1\) such that there exists \(k^{\prime}\) for which \(b_{k^{\prime}}=a_{n^{\prime}}\). There exists a natural number \(k > k^{\prime}\) such that \(b_kL < \frac{\varepsilon}2\). Let \(n_0\) be an integer such that \(a_{n_0}=b_k\). Then for \(n\geq n_0\) we have \[a_nL\leqa_na_{n_0}+
a_{n_0}L < \frac{\varepsilon}2+b_kL < \frac{\varepsilon}2+\frac{\varepsilon}2=\varepsilon.\]
The number \(e\)
Theorem 11 (Number \(e\))
Let \((a_n)_{n=1}^{\infty}\) be the sequence defined as: \[a_n=\left(1+\frac 1n\right)^n.\]
This sequence is convergent.
Remark. The limit of the previous sequence is denoted by \(e\). Its approximate value is \(e\approx 2.718281828\dots\).
We will use the following inequality (Bernoulli’s inequality): For all \(a > 1\) and \(n\in\mathbb N\) we have \[(1+a)^n > 1+na.\] The inequality has an easy proof by induction.
We will consider another sequence \((b_n)_{n=1}^{\infty}\) defined by \(b_n=\left(1+\frac1n\right)^{n+1}\) and prove that:
 (a) The sequence \((a_n)_{n=1}^{\infty}\) is increasing.
 (b) The sequence \((b_n)_{n=1}^{\infty}\) is decreasing.
 (c) For all \(n\in\mathbb N\) we have \(a_n\leq b_n\).

(a) We need to prove that \[\left(1+\frac1n\right)^n\leq \left(1+\frac1{n+1}\right)^{n+1}.\] The last inequality is equivalent to \(\frac{n+1}{n+2}\leq \left(\frac{n(n+1)}{(n+1)^2}\right)^n\). Using Bernoulli’s inequality we get:
\[ \left(\frac{n(n+1)}{(n+1)^2}\right)^n=\left(1\frac1{(n+1)^2}\right)^n\geq 1\frac{n}{(n+1)^2}.\]
Now it remains to prove \(1\frac{n}{(n+1)^2}\geq\frac{n+1}{n+2}=1\frac1{n+2}\) and
this follows from \(\frac1{n+2} > \frac{n}{(n+1)^2}\) which holds because \((n+1)^2=n(n+2)+1\).

(b) We need to prove the inequality \[\left(1+\frac1n\right)^{n+1}\geq \left(1+\frac1{n+1}\right)^{n+2}.\]
The inequality is equivalent to \((n+1)^{2n+3}\geq\left(n(n+2)\right)^{n+1}\cdot(n+2)\), which is equivalent to \[\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+1}\geq\frac{n+2}{n+1}.\]
Bernoulli’s inequality implies: \[\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+1}=\left(1+\frac1{n(n+2)}\right)^{n+1}\geq 1 +\frac{n+1}{n(n+2)}.\]
It remains to establish \(\frac{n+1}{n(n+2)}\geq \frac{1}{n+1}\). The last inequality immediately follows from \((n+1)^2 > n(n+2)\).
(c) The inequality \(a_n\leq b_n\) is obvious because \(\frac{b_n}{a_n}=1+\frac1n > 1\).
Now we have that the sequence \(\left(a_n\right)_{n=1}^{\infty}\) is increasing and bounded above. Each element of the sequence \(\left(b_n\right)_{n=1}^{\infty}\) is an upper bound for \(\left(a_n\right)_{n=1}^{\infty}\). Similarly, the sequence \(\left(b_n\right)_{n=1}^{\infty}\) is decreasing and bounded below. Each element of \(\left(a_n\right)_{n=1}^{\infty}\) is a lower bound for the sequence \(\left(b_n\right)_{n=1}^{\infty}\). Thus both sequences \(\left(a_n\right)_{n=1}^{\infty}\) and \(\left(b_n\right)_{n=1}^{\infty}\) are convergent and the following inequality holds
\[\lim_{n\to\infty}a_n\leq \lim_{n\to\infty}b_n.\]
We will now prove that the limits are equal. We will prove that \(\lim_{n\to\infty}\left(b_na_n\right)=0\). The difference between the terms \(b_n\) and \(a_n\) is
\begin{eqnarray*}
b_na_n&=&\left(1+\frac1n\right)^{n+1}\left(1+\frac1n\right)^n=\left(1+\frac1n\right)^n\cdot \frac1n\\
&=&a_n\cdot\frac1n.
\end{eqnarray*}
Since \(a_n\) is bounded above by \(b_0\), we have \[\leftb_na_n\right\leq b_0\cdot \frac1n.\]
For every \(\varepsilon > 0\), we can choose \(n_0 > \frac{b_0}{\varepsilon}\). If \(n\) is an integer that satisfies \(n\geq n_0\), the it is obvious that \(b_na_n < \varepsilon\).
We have now established that \(\displaystyle\lim_{n\to\infty}(b_na_n)=0\), hence
\[\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty}\left(1+\frac1n\right)^{n+1}.\]
Practice Problems
Problem 1.
Determine whether the sequence \((a_n)_{n=1}^{\infty}\) is convergent. If it is, find its limit. \[a_n=\frac{2n+19n^2+\frac3n}{4n29 n^212}.\]
The sequence is convergent and its limit is \(\frac{19}{29}\) because:
\[\lim_{n\to\infty}\frac{2n+19 n^2+\frac3n}{4n29 n^212}=
\lim_{n\to\infty}\frac{19+\frac{2}{n}+\frac3{n^3}}{29\frac4n\frac{12}{n^2}}.
\]
Let us denote \(b_n=19+\frac{2}{n^3}+\frac3{n^3}\) and \(c_n=29\frac4n\frac{12}{n^2}\). The sequences \((b_n)_{n=1}^{\infty}\) and \((c_n)_{n=1}^{\infty}\) are convergent and their limits are \(19\) and \(29\), respectively. Thus \((a_n)_{n=1}^{\infty}=\left(\frac{b_n}{c_n}\right)_{n=1}^{\infty}\) is convergent and its limit is \(\frac{19}{29}\).
Problem 2.
Determine whether the sequence \((a_n)_{n=1}^{\infty}\) is convergent. If it is, find its limit. \[a_n=\frac{3n^2+\cos n}{4n^225}.\]
The sequence is convergent and its limit is \(\frac{3}{4}\) because:
\[\lim_{n\to\infty}\frac{3n^2+\cos n}{4n^225}=
\lim_{n\to\infty}\frac{3+\frac{\cos n}{n^2}}{4\frac{25}{n^2}}.
\]
Let us denote \(b_n=
3+\frac{\cos n}{n^2}
\) and \(c_n=4\frac{25}{n^2}\). The sequences \((c_n)_{n=1}^{\infty}\) is convergent and its limit is \(4\).
The convergence of the sequence \(b_n\) can be established using the squeeze theorem. Set \(p_n=3\frac1{n^2}\) and \(q_n=3+\frac1{n^2}\). Since \(\cos n\in[1,1]\) we have that \(p_n\leq b_n\leq q_n\). Since \(\displaystyle \lim_{n\to\infty}p_n=\lim_{n\to\infty}q_n=3\) we have \(\displaystyle \lim_{n\to\infty}b_n=3\).
Thus \((a_n)_{n=1}^{\infty}=\left(\frac{b_n}{c_n}\right)_{n=1}^{\infty}\) is convergent and its limit is \(\frac{3}{4}\).
Problem 3.
Determine whether the sequence \((a_n)_{n=1}^{\infty}\) is convergent. If it is, find its limit. \[a_n=\left(1\frac1{3n}\right)^n.\]
We first notice that
\begin{eqnarray*}a_n&=&\left[\left(1\frac1{3n}\right)^{3n}\right]^{\frac1{3}}=\left[\left(\frac{3n1}{3n}\right)^{3n}\right]^{\frac1{3}}=
\frac1{
\left[\left(1+\frac1{3n1}\right)^{3n1}
\right]^{\frac1{3}}
}
\cdot
\frac1{
\left[1+\frac1{3n1}
\right]
^{\frac1{3}
}
}
.
\end{eqnarray*}
Let us introduce the following two sequences \(b_n\) and \(c_n\): \begin{eqnarray*}
b_n&=&\frac1{
\left[\left(1+\frac1{3n1}\right)^{3n1}
\right]^{\frac1{3}}}\\
c_n&=&\frac1{
\left[1+\frac1{3n1}
\right]
^{\frac1{3}
}
}.
\end{eqnarray*}
We will prove that \(\displaystyle \lim_{n\to\infty} b_n\) and \(\displaystyle \lim_{n\to\infty} c_n\) exist. Then the limit of the sequence \((a_n)_{n=1}^{\infty}\) will be product of the previous two limits.
We start by finding \(\displaystyle \lim_{n\to\infty} c_n\):
\(\displaystyle\lim_{n\to\infty} \left(1+\frac1{3n1}\right)=1\), hence \(\displaystyle \lim_{n\to\infty}\left(1+\frac1{3n1}\right)^{\frac1{3}}=1^{\frac13}=1\) and \(\displaystyle \lim_{n\to\infty}c_n=1\).
Let us now find \(\displaystyle \lim_{n\to\infty} b_n\). We first have that \(\displaystyle \lim_{n\to\infty}\left(1+\frac1{3n1}\right)^{3n1}=e\) since this is a subsequence of \(\left(\left(1+\frac1n\right)^n\right)_{n=1}^{\infty}\). Therefore \(\displaystyle \lim_{n\to\infty}
\left[\left(1+\frac1{3n1}\right)^{3n1}\right]^{\frac1{3}}= e^{\frac1{3}}\) and \(\displaystyle \lim_{n\to\infty} b_n=\frac1{e^{\frac1{3}}}\).
We can now conclude that \(\displaystyle\lim_{n\to\infty} a_n=\frac1{e^{\frac1{3}}}\cdot 1=\frac1{e^{\frac1{3}}}\).
Problem 4.
Determine whether the sequence \((a_n)_{n=1}^{\infty}\) is convergent. If it is, find its limit. \[a_n=\frac{5n^2+\cos n}{4n25}.\]
We will prove that the sequence is unbounded, and hence divergent. Assume that \(M > 0\) is arbitrary real number. We will prove that there is a term of the sequence that is greater than \(M\).
Given \(M > 0\), let us pick \(n=\max\{M,6\}\). Then since \(n > 1\) we have \(5n^2+\cos n\geq 5n^21 > 4n^2\). On the other hand, \(4n25\leq 4n\) and \(4n25 > 0\) hence \[\frac{5n^2+\cos n}{4n25} > \frac{4n^2}{4n}=n\geq M.\]
Problem 5.
Determine whether the sequence \((a_n)_{n=1}^{\infty}\) is convergent. If it is, find its limit. \[a_n=1+\frac{\sin n}{n^2}.\]
The convergence of the sequence \(a_n\) can be established using the squeeze theorem. Set \(p_n=1\frac1{n^2}\) and \(q_n=1+\frac1{n^2}\). Since \(\sin n\in[1,1]\) we have that \(p_n\leq a_n\leq q_n\). Since \(\displaystyle \lim_{n\to\infty}p_n=\lim_{n\to\infty}q_n=1\) we have \(\displaystyle \lim_{n\to\infty}a_n=1\).
Thus \((a_n)_{n=1}^{\infty}=\left(\frac{b_n}{c_n}\right)_{n=1}^{\infty}\) is convergent and its limit is \(\frac34\).