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?!
['https://math.stackexchange.com/questions/3870947', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/-1/']
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.

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