Examine the following sequences of numbers $(a_n)$ with $n \in \mathbb{N}$ for the properties of monotonicity and boundedness. As per the title, I need to examine the following sqeuences of numbers $(a_n)$ with $n \in \mathbb{N}$ for the properties of monotonicity and boundedness. These sequences are given via the following functions, which are to be examined:


*

*$a_n = \frac{n^2}{n!}$

*$b_n = \frac{(-2)^{n+1}+3^n}{3^{n+1}+(-2)^n}$

*$c_{n+1} = \sqrt{2+c_n}; c_0 = \sqrt{2}$



Since sequences are a new topic for me, I will refer to the following definitions:


*

*Sequence (mathematics): (here) The resulting ordered set of numbers produced from a function.

*Monotonicity

*Boundedness
Also, here I am avoiding limits as I believe they are being purposely excluded from this task.


Progress so far:


*

*Here I wasn't really sure what to expect, so i just decided to substitute $n = \{0, 1, 2, 3, 4\}$ which produces the set $\{0, 1, 2, \frac{3}{2}, \frac{2}{3}\}$. So from observation, the set of numbers is increasing until the value of $b_2$, but then the set is decreasing. Thus, before any proof, it would appear that the function $a$ is monotonically decreasing, but not strictly. 


To prove this, I write an inequality with $n$ and $n + 1$. 
$
\begin{align*}
\frac{n^2}{n!} &> \frac{(n+1)^2}{(n+1)!} \\
\frac{n^2}{n!} &> \frac{(n+1)^2}{(n+1)n!} \\
n^2 &> \frac{(n+1)^2}{(n+1)} \\
n^2 &> n + 1 \\
\end{align*}
$
By substituting values for n it can be seen that the function is monotonically decreasing from n $\geq 2$. 
To prove boundedness, for the upper bound we observe that $\frac{n^2}{n!} \leq 2$ (supremum). For the lower bound we consider that $n!$ grows faster than $n^2$, and both are always greater than $0$. Thus $n^2/n!$ will tend towards the (lower bound) value of $0$ (infimum). 
Thus, $a_n$ is nonstrictly monotonically decreasing and bounded. 



*Here, the first thing I notice is that the sign of the function changes every time n increases. So according to the definition I would reason that the function is neither increasing nor decreasing, i.e the function is not monotonic.


the second thing that I notice is that the function is composed of powers of (negative) 2 and 3. Since powers of 3 grow faster than powers of 2, I reason that for very large values of n the function will tend towards $\frac{3^n}{3^{n+1}} = \frac{1}{3}$ (the infimum and supremum are the same), i.e $b_n$ is non-monotonic but bounded (This feels very non-rigorous however, is there a better way of handling this question?) 



*This function appears to have been given in recursive notation (as opposed to index notation). The first 3 values are as follows:


\begin{align*}
c_0 &= \sqrt{2} \\
c_1 &= \sqrt{2 + \sqrt{2}} \\
c_2 &= \sqrt{2 + \sqrt{2 + \sqrt{2}}}\\
\end{align*}
Monotonicity: Since more terms are added for each successive recursive call, and the elements are greater than zero, I would reason that the function is monotonically increasing. 
For the boundedness of the function, After plugging a few values into wolfram, it appears that this sequence tends towards 2. 
Can anyone offer some pointers? At this stage I get stuck.
 A: I will answer the 3rd question since I think you did pretty good for the first two.
$$c_{n+1}=f(c_n)$$
with $f(x)=\sqrt{2+x}$ and $c_0=\sqrt 2$


*

*Prove by recurrence that $\forall n \geq 0, \ \sqrt2 \leq \ c_n\leq 2 $:
. $\sqrt2 \leq \ c_0\leq 2 $ is true 
. $$\begin{align}  
  \sqrt2 & \leq \ c_n  & \leq & 2  \\
  2+\sqrt2 &  \leq \ 2+c_n & \leq  & 2+2  \\
  \sqrt{2+\sqrt2} & \leq  \ \sqrt{2+c_n} & \leq  & \sqrt{2+2}    \\
  \sqrt{2} & \leq \ c_{n+1} & \leq & 2 
\end{align}$$ 

*Prove that $\forall x\in [\sqrt 2 , 2] , \ |f'(x)|\leq \lambda < 1$
$$f'(x)=|f'(x)| = \frac{1}{2\sqrt{2+x}} \leq \frac{1}{2\sqrt{2+\sqrt 2}} = \lambda <1 $$

*Prove existence and uniqueness of the fixed point of $f$ i.e: $\exists! \alpha \in [\sqrt 2, 2]$ s.t  $f(\alpha) = \alpha $
$$\begin{align}  x \in [\sqrt 2, 2],\  f(x) =x & \iff \sqrt{2+x}=x \\ & \iff x^2-x-2 =0  & (\mathrm{because \ }x>0) \\ & \iff x=2  & (\mathrm{because \ }x>0)   \end{align}$$

*Prove that $\lim_{n \to \infty} c_n = 2$
$$ c_{n+1} = f(c_n) \iff  \alpha-c_{n+1} = f(\alpha)-f(c_n) $$
f is continuous we can apply the mean value theorem $\exists a_n \in (c_n, \alpha)$ s.t $ f(\alpha) - f(c_n) = f'(a_n)(\alpha - c_n) \iff \alpha - c_{n+1} = f'(a_n)(\alpha - c_n)  $
Yet, $ \forall n\geq 0, \ f'(a_n)\leq \lambda <1$ so $$0< \alpha - c_{n+1} \leq \lambda ^n (\alpha-c_0) \iff \lim_{n \to \infty} c_n  = \alpha = 2$$
A: For the 1st one, you can see that $\frac{n^2}{n!}=\frac{n}{(n-1)!}=\frac{n+1-1}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}\leq1+1=2, \forall n\geq2$ where $a_1=1$ so $a_n\leq2$ for all $n\in\Bbb N$.
Also, $\frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)^2}{(n+1)!}}{\frac{n^2}{n!}}=\frac{(n+1)^2n!}{n^2(n+1)!}=\frac{(n+1)^2}{n^2(n+1)}=\frac{n+1}{n^2}\leq1,\forall n\geq2$ , so $a_n$ is decreasing sequence for all $n\in\Bbb N$ since $a_1=2>\frac{3}{4}=a_2$.
