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?!
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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,$$1=\inf\left\{\frac{n+1}n\,\middle|\,n\in\Bbb N\right\}.$$
**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 $$\lim\_{n\to\infty}\frac{n+1}n=1.$$