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 $\lim\_{n\to \infty} a\_n = \inf a\_n$.
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 $lim\_{n\rightarrow\infty}x\_n=1$