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Limit of Sequence

Sequences
For each number of $1,2,3,\ldots,n,\ldots$, there corresponds unique real number

$\displaystyle a_{1},a_{2},a_{3},\ldots,a_{n},\ldots $

We call this list of numbers sequence and call $a_{1},a_{2},\ldots$ term.
NOTE Once the $n$th term is found, all terms can be derived. Thus we call the $n$th term general term.
Bounded sequence
A sequence $\{a_{n}\}$ is called bounded above if there exists a number $M$ such that $a_{n} \leq M$ for all $n$. A sequence $\{a_{n}\}$ is called bounded below if there exists a number $m$ such that $a_{n} \geq m$ for all $n$. Furthermore, a sequence $\{a_{n}\}$ is called bounded if it is bounded above and bounded below..
NOTE The sequence $\{a_{n}\} = \{\frac{1}{n}\}$ is bounded. For $\frac{1}{n} > 0$ and $\frac{1}{n} < 1$ for all $n$.

Monotonicity
If $a_{n+1} \geq a_{n}$ for all $n$, a sequence $\{a_{n}\}$ is called monotonically increasing. If $a_{n+1} \leq a_{n}$ for all $n$, a sequence $\{a_{n}\}$ is called monotonicall decreasing sequence.

Example 1..14   Determine the sequence $\displaystyle{\{a_{n}\} = \{\frac{n}{n+1}\}}$ is monotonically increasings.

SOLUTION We use the ratio to check to see. Note that $a_{n+1} = \frac{n+1}{n+2}$.

$\displaystyle \frac{a_{n+1}}{a_{n}} = \frac{n+1}{n+2}\cdot \frac{n+1}{n} = \frac{n^{2}+2n+1}{n^{2}+2} > 1$

Thus $a_{n+1} \geq a_{n}$ and $\{a_{n}\}$ is monotonically increasing $\ \blacksquare$

Exercise 1..14   Determine whether the following sequence is increasing or decreasing.

$\displaystyle a_{1} = 1, a_{n+1} = \sqrt{3a_{n}}$

SOLUTION Since $a_{1} = 1, a_{2} = \sqrt{3a_{1}} = \sqrt{3}$ , we have $a_{1} < a_{2}$. Now use mathematical induction. 1.5

For $n=1$, $a_{n} < a_{n+1}$ is true. Now assume that $a_{k} < a_{k+1}$ is true. Then we have

$\displaystyle a_{k+1} = \sqrt{3a_{k}} < \sqrt{3a_{k+1}} = a_{k+2} $

Thus by mathematical induction, $a_{n} < a_{n+1}$ for all $n$ $\ \blacksquare$

Limit of sequence
As $n$ approaches $\infty$, $a_{n}$ approaches $a$. Then we write

$\displaystyle \lim_{n \rightarrow \infty}a_{n} = a $

We say the sequence $\{a_{n}\}$ converges to $a$ and call this $a$ limit.
NOTE When a sequence converges to $a$, $a$ has to be a real number. Thus we can not use $+\infty$ or $-\infty$ for $a$. When a sequence $\{a_{n}\}$ does not converge, we say $\{a_{n}\}$ diverges.

Limit properties

Theorem 1..1   Suppose that $\displaystyle{\lim_{n \rightarrow \infty}a_{n} = a, \lim_{n \rightarrow \infty}b_{n} = b}$ and $c$ is constant. Then we have the followings:

$\displaystyle{1. \ \lim_{n \rightarrow \infty}(a_{n} \pm b_{n})= a \pm b}$

$\displaystyle{2. \ \lim_{n \rightarrow \infty}(ca_{n}) = ca \ }$

$\displaystyle{3. \ \lim_{n \rightarrow \infty}a_{n}b_{n} = ab}$

$\displaystyle{4. \ \lim_{n \rightarrow \infty}\frac{a_{n}}{b_{n}} = \frac{a}{b} \ \mbox{\rm provided}\ b \neq 0}$

NOTE When the limit exits, four arithmetic operations hold. Express 1. by

$\displaystyle \lim_{n \rightarrow \infty}(a_{n} \pm b_{n}) = \lim_{n \rightarrow \infty}a_{n} \pm \lim_{n \rightarrow \infty} b_{n}$

Then we can memorize by saying "the limit of a sum is the sum of limits".

Limit properties

Theorem 1..2   Suppose that $\displaystyle{\lim_{n \to \infty}a_{n} = \infty, \lim_{n \to \infty}b_{n} = b > 0}$. Then we have the followings:

$\displaystyle{1. \ \lim_{n \to \infty}\frac{a_{n}}{b_{n}} = \infty}$

$\displaystyle{2. \ \lim_{n \to \infty}\frac{b_{n}}{a_{n}} = 0}$

NOTE Suppose that the denominator of the sequence approaches some positive constant as the numerator approaches $\infty$, Then the sequence gets larger without bound. Thus the sequence diverges. Suppose next that the denominator of the sequence approaches $\infty$ as the numerator approaches a positive constant. Then the limit of the sequence is 0.

Example 1..15   Find the limit of the following.

$\displaystyle \{a_{n}\} = \{\frac{3n^2 - 5n}{5n^2 + 2n - 6}\}$

SOLUTION As $n \rightarrow \infty$, $\displaystyle{3n^2 - 5n = n^2 (3 - \frac{5}{n})}$ $\longrightarrow \infty$ and $5n^2 + 2n - 6 =$ $\displaystyle{n^2 (5 + \frac{2}{n} - \frac{6}{n^2}) \longrightarrow \infty}$. Then we factor by taking out $n^2$ to have

$\displaystyle \lim_{n \rightarrow \infty}\frac{3n^2 - 5n}{5n^2 + 2n - 6}$ $\displaystyle =$ $\displaystyle \lim_{n \rightarrow \infty}\frac{n^2(3 - 5/n)}{n^2(5 + 3/n - 6/n^2)}$  
  $\displaystyle =$ $\displaystyle \lim_{n \rightarrow \infty}\frac{3 - 5/n}{5 + 3/n - 6/n^2} = \frac{3}{5}\ensuremath{\ \blacksquare}$  

Exercise 1..15   Find the limit of the sequence $\displaystyle{\{\sqrt{n+1} - \sqrt{n}\}}$

SOLUTION
$\displaystyle \lim_{n \to \infty}(\sqrt{n+1} - \sqrt{n})$ $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}}$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\frac{n+1 -n}{\sqrt{n+1} + \sqrt{n}} = \lim_{n \to \infty}\frac{1}{\sqrt{n+1} + \sqrt{n}}= 0 \ensuremath{\ \blacksquare}$  

To find the limit, the theorem above is not enough. For example, consider $\displaystyle{\lim_{n \rightarrow \infty}\frac{\sin{n\theta}}{n}}$ as $\theta \neq 0$. In this problem, we can not obtain the limit by using the theorem 1.1 and the theorem 1.2. To find the limit of $\sin{n\theta}$, it is useful to use the following theorem.

Squeezing theorem

Theorem 1..3   If there exists a number $n_0$ so that $a_{n} < c_{n} < b_{n}$ is true for all $n > n_0$ and

$\displaystyle \lim_{n \rightarrow \infty}a_{n} = \lim_{n \rightarrow \infty}b_{n} = a $

Then $\lim_{n \rightarrow \infty}c_{n} = a$.

Proof. Since $a_{n} < c_{n} < b_{n}$, we have either $0 \leq \vert c_{n} - a\vert < \vert a_{n} - a\vert$ or $0 \leq \vert c_{n} - a\vert <\vert b_{n} - a\vert$. Note that

$\displaystyle \lim_{n \rightarrow \infty}a_{n} = \lim_{n \rightarrow \infty}b_{n} = a $

Thus $\lim_{n \rightarrow \infty}c_{n} = a$ $\ \blacksquare$

Example 1..16   For $\theta \neq 0$, find the limit of the following.

$\displaystyle \{a_{n}\} = \{\frac{1}{n}\sin{n\theta}\}$

SOLUTION Since $\vert\sin{n \theta}\vert \leq 1$, we have

$\displaystyle 0 \leq \vert\frac{\sin{n \theta}}{n}\vert \leq \frac{1}{n} $

for all $n, \theta$. Thus we can sandwich $\displaystyle{\vert\frac{\sin{n \theta}}{n}\vert}$ using 0 and $\displaystyle{\frac{1}{n}}$. Now taking the limit of 0 and $\displaystyle{\frac{1}{n}}$, we obtain

$\displaystyle \lim_{n \rightarrow \infty} 0 = 0, \ \lim_{n \rightarrow \infty} \frac{1}{n} = 0. $

Thus by the squeezing theorem,

$\displaystyle \lim_{n \rightarrow \infty}\vert\frac{\sin{n \theta}}{n}\vert = 0. $

Note that

$\displaystyle \vert\frac{\sin{n \theta}}{n} - 0\vert = \vert\vert\frac{\sin{n \theta}}{n}\vert - 0 \vert $

we have

$\displaystyle \lim_{n \rightarrow \infty}\frac{\sin{n \theta}}{n} = 0\ensuremath{\ \blacksquare}$

Exercise 1..16   Find the following limit.
1. $\displaystyle{\lim_{n \to \infty}\frac{(-2)^n}{1 + 3^n}}$2. $\displaystyle{\lim_{n \to \infty}\frac{5^{n+1} + 4^n}{3^{n-1} - 5^n}}$

SOLUTION 1. $\displaystyle{\lim_{n \to \infty}\frac{(-2)^n}{1 + 3^n} = \lim_{n \to \infty}\f... ...rac{-2}{3}\big)^n}{\big(\frac{1}{3}\big)^n + 1} = 0\ensuremath{\ \blacksquare}}$ 2. $\displaystyle{\lim_{n \to \infty}\frac{5^{n+1} + 4^n}{3^{n-1} - 5^n} = \lim_{n ... ...big(\frac{3}{5}\big)^{n-1} - 1} = \frac{5}{-1} = -5\ensuremath{\ \blacksquare}}$

The limit of $\{r^n\}$ is a base of convergence and divergence.

Bernoulli inequality
For $x > 0$ and $n > 1$,

$\displaystyle x^{n} \geq n(x - 1) + 1 $

Proof $x^{n}-1 = (x-1)(x^{n-1}+x^{n-2}+\cdots+x+1) $.
For $x > 1$, we have $(x-1) > 0$ and $x^{n-1}+x^{n-2}+\cdots+x+1 > 1 + 1 + \cdots + 1 = n$
For $x = 1$, we have $(x-1) = 0$ and $x^{n-1}+x^{n-2}+\cdots+x+1 = 1 + 1 + \cdots + 1 = n$
For $0 < x < 1$, we have $(x-1) < 0$ and $x^{n-1}+x^{n-2}+\cdots+x+1 < 1 + 1+ \cdots + 1 = n$. Thus,

$\displaystyle x^{n}-1 = (x-1)(x^{n-1}+x^{n-2}+\cdots+x+1) \geq n(x-1) \ensuremath{\ \blacksquare}$

Example 1..17   For $\displaystyle{a_{n} = \frac{r^{n+1}}{1 + r^n}\ (r \neq -1)}$, find $\lim_{n \to \infty}a_{n}$.

SOLUTION We first find the limit of $\{r^n\}$.

$\displaystyle \lim_{n \rightarrow \infty}r^{n} = \left\{\begin{array}{ll} \inft... ...\ 0 & (\vert r\vert < 1)\\ \mbox{divergence} & (r \leq -1) \end{array}\right.$

Now for $r = 1$, $a_{n} = \frac{1}{2}$ and $\displaystyle{\lim_{n \to \infty}a _{n} = \frac{1}{2}}$. For $\vert r\vert < 1$, $\lim_{n \to \infty}r^n = \lim_{n \to \infty}r^{n+1} = 0$ and $\lim_{n \to \infty}a_{n} = 0$.

For $r > 1, r < -1$,

$\displaystyle \lim_{n \to \infty}a_{n} = \lim_{n \to \infty}\frac{r}{\left(\frac{1}{r}\right)^n + 1} = r\ensuremath{\ \blacksquare}$

Exercise 1..17   Find the sequence $\displaystyle{\{a_{n}\} = \{(1 + \frac{1}{n})^{n}\}}$ is monotonically increasing.

SOLUTION Since $\displaystyle{\{a_{n}\} = \{(1 + \frac{1}{n})^{n}\}}$, we have

$\displaystyle a_{n} = (1 + \frac{1}{n})^{n}= (\frac{n+1}{n})^{n},\ a_{n-1} = (1+\frac{1}{n-1})^{n-1} = (\frac{n}{n-1})^{n-1}.$

Then,
$\displaystyle \frac{a_{n}}{a_{n-1}}$ $\displaystyle =$ $\displaystyle (\frac{n+1}{n})^{n}(\frac{n-1}{n})^{n-1} = \frac{n}{n-1}[(\frac{n+1}{n})(\frac{n-1}{n})]^n$  
  $\displaystyle =$ $\displaystyle \frac{n}{n-1}(\frac{n^{2}-1}{n^{2}})^{n} \geq \frac{n}{n-1}(1 + n(\frac{n^{2}-1}{n^{2}} - 1))$  
  $\displaystyle =$ $\displaystyle \frac{n}{n-1}(\frac{n-1}{n}) = 1.$  

Thus, $a_{n} \geq a_{n-1}$ and $\{a_{n}\}$ is monotonically increasing $\ \blacksquare$
Monotone convergence theorem

Theorem 1..4   Every increasing sequence that is bounded above converges. Every decreasing sequence that is bounded below converges..

NOTE If a sequence $\{a_{n}\}$ is monotonically increasing and bounded above, then there exists a number for which the sequence $\{a_{n}\}$ can not become greater than that number. Among all those numbers, we let the least number be $s$. Then the difference between $s$ and $a_{n}$ becomes small. Thus the sequence $\{a_{n}\}$ converges.

Example 1..18   Given that $\displaystyle{a_{n+1} = \sqrt{3a_{n}}, a_{1} = 1}$. Determine the sequence $\{a_{n}\}$ converges or diverges..

SOLUTION We show by induction that the sequence $\{a_{n}\}$ converges. By Exercise1.14, we know $\{a_{n}\}$ is monotonically increasing sequence. So, we need to show the sequence is bounded above. We use mathematical induction on $n$.

Since $a_{1} = 1 < \sqrt{3} < 3$, it is true for $n=1$.

Assume that $a_{k} < 3$. Then $a_{k+1} = \sqrt{3a_{k}} < \sqrt{3\cdot3} = 3$. Thus for all $n$, $a_{n} < 3$ and $\{a_{n}\}$ is bounded above increasing sequence. Therefore, $\{a_{n}\}$ converges $\ \blacksquare$

Exercise 1..18   Given $\displaystyle{a_{n} = (1 + \frac{1}{n})^n}$. Determine the sequence $\{a_{n}\}$ converges or diverges.

SOLUTION We have shown in Exercise1.17 that $\displaystyle{\{(1 + \frac{1}{n})^n\}}$ is monotonically increasing. Thus we need to show the sequence is bounded above.

Expand $\displaystyle{(1 + \frac{1}{n})^n}$ 1.6using binomial theorem,

$\displaystyle a_{n}$ $\displaystyle =$ $\displaystyle \left(1 + \frac{1}{n}\right)^{n} = \binom{n}{0} \left(\frac{1}{n}... ...binom{n}{1} \left(\frac{1}{n} \right) + \binom{n}{2} \left(\frac{1}{n}\right)^2$  
  $\displaystyle +$ $\displaystyle \cdots + \binom{n}{n} \left(\frac{1}{n}\right)^n$  
  $\displaystyle =$ $\displaystyle 1 + n\cdot \frac{1}{n} + \frac{n(n-1)}{2!}\left(\frac{1}{n}\right)^2 + \cdots + \left(\frac{1}{n}\right)^n$  
  $\displaystyle <$ $\displaystyle 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots < 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots \leq 3.$  

Thus, the sequence $\{a_{n}\}$ is bounded above increasing sequnce and converges $\ \blacksquare$

Euler e

$\displaystyle e = \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n $

Example 1..19   Find the limit of the following.

$\displaystyle \lim_{n \to \infty}(1 + \frac{2}{n})^{n}$

SOLUTION Put $\frac{2}{n} = \frac{1}{m}$. Then $m = \frac{n}{2}$ and $n \to \infty$ implies that $m \to \infty$. Thus,

$\displaystyle \lim_{n \to \infty}(1 + \frac{2}{n})^{n} = \lim_{m \to \infty}(1 ... ...o \infty}\big((1 + \frac{1}{m})^{m}\big)^{2} = e^{2}\ensuremath{\ \blacksquare}$

Exercise 1..19   Find the limit of the following.

$\displaystyle \lim_{n \to \infty}(1 - \frac{1}{n})^{n}$

SOLUTION Since $1 - \frac{1}{n} = \frac{n-1}{n} = \frac{1}{\frac{n}{n-1}} = \frac{1}{1 + \frac{1}{n-1}}$, we have
$\displaystyle \lim_{n \to \infty}(1 - \frac{1}{n})^{n}$ $\displaystyle =$ $\displaystyle \lim_{n \to \infty}(\frac{1}{1 + \frac{1}{n-1}})^{n} = \lim_{n \to \infty}\frac{1}{(1 + \frac{1}{n-1})^{n-1}(1+ \frac{1}{n-1})}$  
  $\displaystyle =$ $\displaystyle \frac{1}{e}\ensuremath{\ \blacksquare}$  

Limit ratio test

NOTE Since $\lim_{n \rightarrow \infty} \vert\frac{a_{n+1}}{a_{n}}\vert = r$, for all $n$ such that $n \geq N$, we have
$\displaystyle \vert a_{N+1}\vert$ $\displaystyle \leq$ $\displaystyle r\vert a_{N}\vert$  
$\displaystyle \vert a_{N+2}\vert$ $\displaystyle \leq$ $\displaystyle r\vert a_{N+1}\vert \leq r^2\vert a_{N}\vert$  
  $\displaystyle \vdots$    
$\displaystyle \vert a_{N+n}\vert$ $\displaystyle \leq$ $\displaystyle r\vert a_{N+n-1}\vert \leq r^{n}\vert a_{N}\vert$  

Noting that $0 \leq r < 1$, we have $\lim_{n \to \infty} r^{n} = 0$. Thus, $\lim_{n \rightarrow \infty} a_{n} = 0.$

Example 1..20   Find the limit of the sequence $a_{n+1} = \sqrt{3a_{n}}, \ a_{1} = 1$

SOLUTION Note that if $\{a_{n}\}$ converges to $\alpha$, then $\{a_{n+1}\}$ converges to $\alpha$. Thus we have

$\displaystyle \alpha = \sqrt{3\alpha} $

Solve this to get $\alpha = 0, 3$. Since $a_{1} = 1$, it is impossible to have $\alpha = 0$. So we have $\alpha = 3$. Note this is not the end of proof. We have to show $\lim_{n \to \infty}a_{n} = 3$. By the limit ratio test, it is enough to show $\lim_{n \to \infty}\vert\frac{a_{n+1} - 3}{a_{n} - 3}\vert < 1$.
$\displaystyle \lim_{n \to \infty}\vert\frac{a_{n+1} - 3}{a_{n} - 3}\vert$ $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\vert\frac{\sqrt{3a_{n}} + 3}{a_{n} - 3}\vert = \lim_{n \to \infty}\vert\frac{3a_{n}- 9}{(\sqrt{3a_{n}} + 3)(a_{n} - 3)}\vert$  
  $\displaystyle =$ $\displaystyle \lim_{n \to \infty}\vert\frac{3(a_{n}- 3)}{(\sqrt{3a_{n}} + 3)(a_{n} - 3)}\vert = \lim_{n \to \infty}\vert\frac{3}{\sqrt{3a_{n}} + 3}\vert = 0 < 1.$  

Thus, $\lim_{n \to \infty}a_{n} = 3$ $\ \blacksquare$

Exercise 1..20   Find the limit the following.

$\displaystyle \lim_{n \to \infty} \frac{2^{n}}{n!}$

SOLUTION Set $a_{n} = \frac{2^{n}}{n!}$. Then

$\displaystyle \frac{a_{n+1}}{a_{n}} = \frac{2^{n+1}}{(n+1)!}\cdot \frac{n!}{2^n} = \frac{2}{n+1}$

and

$\displaystyle \lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \lim_{n \to \infty} \frac{2}{n+1} = 0.$

Thus by the limit ratio test, $\displaystyle{\lim_{n \to \infty} \frac{2^{n}}{n!} = 0\ensuremath{\ \blacksquare}}$

Exercise A


1.
Find the limit of the following sequences:?D

(a) $\displaystyle{\{a_{n}\} = \{\sqrt{n}\}}$ (b) $\displaystyle{\{a_{n}\} = \{\frac{n+1}{n^{2}}\}}$ (c) $\displaystyle{\{a_{n}\} = \{\frac{n^{2}}{n + 1}\}}$

(d) $\displaystyle{\{a_{n}\} = \{\frac{2^{n} - 1}{2^{n}}\}}$ (e) $\displaystyle{\{a_{n}\} = \{\frac{1}{n} - \frac{1}{n+1}\}}$

2.
Determine the following sequences are bounded or not?DDetermine also the following sequences are increasing or decreasing.

(a) $\displaystyle{\{a_{n}\} = \{\frac{2}{n}\}}$ (b) $\displaystyle{\{a_{n}\} = \{\sqrt{4 - \frac{1}{n}}\}}$

3.
Find the general term $\{a_{n}\}$ of the following sequences:

(a) $\displaystyle{a_{1} = 1, a_{n+1} = \frac{1}{n+1}a_{n}, \ n \geq 1}$ (b) $\displaystyle{a_{1} = 1, a_{n+1} = a_{n} + 2, \ n \geq 1}$

(c) $\displaystyle{a_{1} = 1, a_{n+1} = a_{n} + 2n + 1, \ n \geq 1}$

4.
Determine the following sequences converge or not. If it converges, find the limit?D

(a) $\displaystyle{a_{1} = 1, a_{n+1} = \frac{1}{e}a_{n}, \ n \geq 1}$ (b) $\displaystyle{a_{1} = 1, a_{n+1} = 2^{n+1}a_{n}}$

(c) $\displaystyle{a_{1} = 1, a_{n+1} = \frac{n}{n+1}a_{n}}$

Exercise B


1.
Find the limit of the following sequences:

(a) $\displaystyle{\{a_{n}\} = \{n^4 - 3n^3\}}$ (b) $\displaystyle{\{a_{n}\} = \{\frac{3n^{2}+5}{4n^{3} - 1}\}}$ (c) $\displaystyle{\{a_{n}\} = \{\frac{1 - n}{n - \sqrt{n}}\}}$

(d) $\displaystyle{\{a_{n}\} = \{\frac{n(n+2)}{n+1} - \frac{n^{3}}{n^{2}+1}\}}$ (e) $\displaystyle{\{a_{n}\} = \{\sqrt{n+1} - \sqrt{n}\}}$

2.
Show $\displaystyle{\lim_{n \rightarrow \infty}\sqrt[n]{a} = 1}$ for $a > 0$?D
3.
Using $\displaystyle{\lim_{n \rightarrow \infty}\sqrt[n]{a} = 1}$, find the limit of the followings:

(a) $\displaystyle{\{ (a^n + b^n)^{\frac{1}{n}}\}}$ for $a > b > 0$. (b) $\displaystyle{\{a_{n}\} = \{(1+2^{n}+3^{n})^{\frac{1}{n}}\}}$

4.
Determine the following sequences converge or not?D

(a) $\displaystyle{2,2^{2},2^{3},\cdots,2^{n},\cdots}$(b) $a_{n}$ is an nth digit approximation of $\sqrt{2}$ ?D

2.
Find the limit of the following sequences $\{a_{n}\}$?D

(a) $\displaystyle{a_{1} = 1, a_{n+1} = \sqrt{3a_{n} + 4}}$ (b) $\displaystyle{a_{1} = 1, a_{2} = 2, a_{n+2} = \sqrt{a_{n+1}a_{n}}}$

3.
Find the limit of the following sequences:

(a) $\displaystyle{a_{n} = (1 - \frac{1}{n^2})^n}$ (b) $\displaystyle{a_n = (1 + \frac{2}{n})^n}$ (c) $\displaystyle{a_n = \frac{2^n}{n!}}$ (Corollary[*])

(d) $\displaystyle{a_n = \frac{n!}{n^n}}$



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