To show a sequence is bounded, monotone and to find its limit I am new to analysis and following is the question:
Show that the sequence $\frac{n+1}{n}$ is monotone, bounded and find its limit.
The way I approached it is the following:
To show that it is monotone,
We can write the sequence as $a_n = 1 + \frac{1}{n}$. Since $n_{2} > n_{1}$, we have that $\frac{1}{n_2}<\frac{1}{n_1} $. And hence $1 + \frac{1}{n_{1}} > 1 + \frac{1}{n_{2}}$. So this shows that the sequence is monotonically decreasing.
Question 1: With analysis I never know if my argument is complete, so is it complete here? Am I missing something?
To show that it is bounded,
We know that since $n\in \mathbb{N}$, we have that $0 < \frac{1}{n} \le 1$, and so $1 < 1+\frac{1}{n} \le 2$. Hence it is bounded.
Question 2: Another analysis question, how do I even know that I am not using things that have not been defined yet? Like, have I taken things for granted in my proof above?
Finally, to find the limit,
Can we just say that since this is a monotonically decreasing sequence, that is also bounded we can say that:
$\lim_{n\rightarrow \infty} x_{n} = inf$ ${x_{n} : n \in \mathbb{N}}$, we can say that the limit in this case would be 1?
Question 3:  I feel like this is not enough, and we would still have to show officially that 1 is the infimum of this sequence, which I am not sure how I can prove without saying that it makes intuitive sense for me? So if someone could tell me what the official proof of this part would be that would be great.
Final Question: Is what I have so far correct or have I made any assumptions that one should not make while solving analysis questions?!
 A: Answers to 1 and 2 are fine.
But you ask a really good question about how do you know what you can accept as obvious or not.
You should have early one spent half a lesson (or told to read) axioms and definitions of the rational/real field.
You therefore have Axiom: if $a < b$ the $a+m < b+m$ for all $m$, and if $c > 0$ than $ac < bc$. And  from there you have a proposition that if $1 < k \iff 0< \frac 1k < 1$ (Pf: if $1 < k$ and $\frac 1k \ge 1$ we'd have the contradictory $\frac 1kk \ge 1\cdot k$.  If $\frac 1k \le 0$ then $1=\frac 1kk<0\cdot k = 0$.  [But note we also have to prove $0\cdot k = 0$, and $1 > 0$ which... well, they should be excercises under your belt]
But once you get through the lesson (which slogs a heck of a lot of stuff; so much stuff that if you  actually gave it the time you think it requires you'd never get to lesson 2) you can assume all basic "facts" about numbers.
And .... well, rule of thumb.  Feel free to say For all $n > 1$ we have $0 < \frac 1n \le 1$ without justification.  BUT be prepared to back it up if you are asked.
Question 3:
Yes  you do have to prove that $\lim_{n\to \infty} \frac {n+1}n = 1$.  But once you see that $\frac {n+1}n =1+\frac 1n$ that's easy with a $N$ epsilon proof:
$|\frac {n+1}n -1|=|(1+\frac 1n) - 1| = |\frac 1n| = \frac 1n < \epsilon \iff$
$\frac 1n < \epsilon \iff$
$n > \frac 1\epsilon$.
So by the definition of $\lim_{n\to \infty} a_n=L$ we have that there exists an $N: = \frac 1\epsilon$ so that $n > N\implies |\frac {n+1}n -1| < \epsilon$.  Thus we have proven $\lim_{n\to \infty} \frac {n+1}n = 1$.
BUT.....  again.... once you prove something once you can assume it is known forever.
You have probably already proven 1) If $\lim a_x =L$ then $\lim (a_x + c) = L +c$ for a constant $c$[$*$] and you have probably already proven 2) $\lim_{n\to \infty} \frac 1n =0$[$**$].
If so, you can just state:  As $\frac {n+1}n$ is bounded below and is monotonic decreasing the limit exists[$***$] and so $\lim_{n\to \infty} \frac {n+1}n = \lim_{n\to \infty}(1 +\frac 1n) = 1+\lim_{n\to \infty} \frac 1n = 1+0 =1$.
That's it.
=======
[$*$]As $|(c+ a_x) -(c+L)| = |a_x - L|$ so $|a_x -L| < \epsilon \iff |(c+ a_x) -(c+L)|<\epsilon$ so $\lim a_x = L \iff \lim (c+a_x) = L + c$.
[$**$]ANd $|\frac 1n -0| = \frac n < \epsilon \iff n > \frac 1{\epsilon}$ so $\lim_{n\to \infty}\frac 1n = 0$.
[$***$] Actually this needs to be justified, but presumably you already have.  It is the basic property of Real Numbers that if a set is bounded above or below then the $\sup$ or $\inf$ exists.  If a set if monotonic and bounded below then. $\lim_{n\to \infty} a_n$ exists and that it must be equal to $\inf a_n$.  This is because for any $\epsilon >0$ then $\inf a_n + \epsilon$ is not a lower bound and there a $N$ so that $\inf a_n \le a_N < \inf a_n + \epsilon$.  And as $a_n$ is monotonic decrease all $k > N$ are such that $\inf a_n \le a_k < a_N < \inf a_n + \epsilon$.  So $|(\inf a_n)-a_k| < \epsilon$.  SO $\lim_{n\to \infty} a_n = \inf a_n$.
A: What you did is just fine. Of course, we are not aware of which theorems you can use. Yes, there is a theorem that says that a bounded and decreasing sequence always converges and that its limit is then the infimum of the set of its terms. And, yes, in order to use it, you will have to prove that$$\inf\left\{\frac{n+1}n\,\middle|\,n\in\Bbb N\right\}=1,$$but that's easy:

*

*for each $n\in\Bbb N$, $1\leqslant1+\frac1n=\frac{n+1}n$, and therefore $1$ is a lower bound of the set $\inf\left\{\frac{n+1}n\,\middle|\,n\in\Bbb N\right\}$;

*if $a>1$, then, by the Archimedian property, there is a $n\in\Bbb N$ such that $n>\frac1{a-1}$. But then $a>1+\frac1n=\frac{n+1}n$, and therefore $a$ is not a lower bound of $\inf\left\{\frac{n+1}n\,\middle|\,n\in\Bbb N\right\}$.

So, $1$ is the greatest lower bound of that set; in other words,$$1=\inf\left\{\frac{n+1}n\,\middle|\,n\in\Bbb N\right\}.$$
A: Your monotonicity and boundedness proofs are very fine.
For the limit existence you can  use the theorem "an increasing (resp. decreasing) sequence upper (resp. lower) bounded is convergent".
But in this case $|x_n-1|=\frac 1n\to 0$ is quite obvious and you do not need much more.
A: You did the proofs for monotonicity and boundedness totally fine.
For the third proof, you can  use the Monotone Convergence Theorem which states that if a sequence is a bounded, decreasing sequence, then the limit of the sequence exists at the greatest lower bound (or infimum) of the sequence.
In other words, if $<x_n>$ is a monotonically decreasing sequence, then $lim_{n\rightarrow\infty}x_n\rightarrow inf$ $x_n, n\in\mathbb{N}$
You've already proved that $x_n=\frac{n+1}{n}=1+\frac{1}{n}$ is a monotonically decreasing sequence and $1\le 1+\frac{1}{n}\le 2$. So the sequence converges to the $inf$ $x_n$.
Now to find the infimum of the sequence $<x_n>$:
Let $r>1$ for any $r\in\mathbb{R}$. Then, $r-1>0$.
Upon applying the Archimedean Property, we will find an $n_0\in\mathbb{N}$ such that, $\frac{1}{r-1}<n_0\Rightarrow r>1+\frac{1}{n_0}$.
Now, as $r$ is arbitrary, thus we have that any $r>1$ is not a lower bound of the sequence $<x_n>$. Thus, the greatest upper bound of the sequence is $1$, since the sequence has a lower bound as $1$.
So, $inf$ $x_n=1$.
Thus, you get that the sequence has the limit as $1$ by the Monotone Convergence Theorem.
There's another easy and interesting way to show that $x_n\rightarrow 1$:
I'll use the result that if $a_n$ and $b_n$ are two sequences such that $a_n\rightarrow a$ and $b_n\rightarrow b$ for some $a,b\in\mathbb{R}$, then $a_n+b_n\rightarrow a+b$.
Simply write $<x_n>=1+\frac{1}{n}$ as the sum of two individual sequences $a_n=1\forall n\in\mathbb{N}$ and $b_n=\frac{1}{n}\forall n\in\mathbb{N}$. Now we observe that $a_n\rightarrow1=a$ and $b_n\rightarrow0=b$.
Thus, $x_n=a_n+b_n\rightarrow a+b=1+0=1 \Rightarrow x_n\rightarrow1$.
This completes the proof that $lim_{n\rightarrow\infty}x_n=1$
A: Monotone:
$$n<m\implies \frac1n>\frac1m\implies\frac1n+1>\frac1m+1$$
is sufficiently formal.
Bounded:
$$n> 0\implies \frac1n>0\implies \frac1n+1>1$$
is sufficiently formal.
Limit:
It is pretty obvious that $\dfrac1n+1$ can be made as close as you want to $1$. This intuition can be validated formally by
$$\forall\epsilon>0:n>N=\frac1\epsilon\implies\left|\frac1n+1-1\right|<\epsilon,$$ which constructively proves that $\exists N$, and $$\lim_{n\to\infty}\frac{n+1}n=1.$$
