Proving sequence convergence

I'm pretty confused...I understand bits and pieces but not how it all comes together...I would appreciate some help, either a written out example (you can make up one) and/or comments on how to fix my partial answers below. Thanks~

1. Using only the definition 4.1.2 prove the following:

"Definition 4.1.2 A sequence $(s_n)$ is said to converge to the real number s provided that for every $ε>0$ there exists a natural number $N$ such that for all $n∈N$, $n \ge\ N$ implies that $|s_n-s|<ε$. If $(s_n)$ converges to s, then s is called the limit of the sequence $(s_n)$ and we write $lim_{n\to \!\, \infty \!\ }s_n=s$. If a sequence does not converge to a real number, it is said to diverge."

I understand the meaning of convergence and divergence but I am a little bit unsure about using this definition to prove the statements. Specifically, I am wondering what ε represents?

a. For any real number k, $lim_{n\to \!\, \infty \!\ }(k/n)=0$

Answer in book: "Given $ε>0$, let $N=\frac{|k|}{ε}$"

Here (I think) I understand the first part, we assume $ε>0$ according to the definition. Then...$s_n=lim_{n\to \!\, \infty \!\ }(k/n)$ and $n \ge\ N$ implies $|s_n-s|<ε$ so in this case $s=0$, so $|s_n|<ε$. Since $n \ge\ N$, $k/n \le\ k/N$. Then $|s_n-s|<ε$ so $|k|/n<ε \le\ |k|/N$ so we let $N=|k|/ε$.

I am especially unsure about the last step(s), if you could help me out with the logic here I would appreciate it.

b. For any real number k>0, $lim_{k\to \!\, \infty \!\ }(\frac{1}{n^k})=0$

Given $ε>0$, let $s_n=lim_{k\to \!\, \infty \!\ }(\frac{1}{n^k})$. Then for all $n∈N$, $n \ge\ N$ implies that $|s_n-s|<ε$ so $|s_n|<ε$. Then since $n \ge\ N$, we have $n^k \ge\ N^k$ and $\frac{1}{n^k} \le\ \frac{1}{N^k}$. And I am a little bit confused about where to continue from here...

Answer in book: "Given $ε>0$, let $N=(\frac{1}{ε})^{1/k}$. Then $n \ge\ N$ implies $n^k>1/ε$, so $1/n^k<ε$"

c. $lim\frac{4n+1}{n+3}=4$

Given $ε>0$, let $s_n=lim\frac{4n+1}{n+3}=4$. Then there exists a natural number $N$ such that for all $n∈N$, $n \ge\ N$ implies that $|s_n-s|<ε$, so |s_n-4|<ε. Since $n \ge\ N$, $\frac{4n+1}{n+3} \ge\ \frac{4N+1}{N+3}$. Because |s_n-4|<ε then |\frac{4n+1}{n+3}-4|=|\frac{4n-3}{n+3}...and then I am confused from there.|

(I will attempt these once I clear up the first 3)

d. $lim\frac{sin}{n}=0$

e. $lim\frac{n+3}{n^2-13}=0$

f. $lim\frac{n+2}{n^2+n-3}=0$

I will try to help you as best as I can.

I recall the definition:

We say that a sequence $(x_n)_{n\in\mathbb{N}}$ converges to a real number $x$ if for all $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n > N$, $|x_n - x| < \varepsilon$. Formally,
\begin{gather*} \forall \varepsilon > 0, \exists N \in \mathbb{N} : |x_n - x| < \varepsilon, \ \forall n > N. \end{gather*}

The first thing to notice about this definition is that the inequalities can be strict or not (except for $\varepsilon$). If you do not understand this fact, just tell me, but I think it is not so difficult for you to see why.

As you have understood, you must fix an $\varepsilon > 0$ and choose (or find) an integer $N$ such that the condition $|x_n - x| < \varepsilon$ holds for all integers $n > N$.

The $\varepsilon$ is arbitrary because you would that your sequence $(x_n)$ becomes as close as possible of $x$. And this is what the definition says. If you make a picture, it means that for every level $\varepsilon > 0$, after a certain integer $N$, the distance from $x_n$ to $x$ is smaller than $\varepsilon$, i.e. for all $n > N$, $x_n \in ]x-\varepsilon , x + \varepsilon[$. So logically, the smaller the $\varepsilon$, the bigger the integer $N$ to find. That is in some ways the principle of the limit: "To be as close as possible of a number as $n \rightarrow \infty \ (\text{that is }\forall n > N)$".

Note that the definition is especially useful for proving theoretical results, because for using it, you have to know first the value of the limit. But in many particular cases, we have some calculation techniques and tricks to find the limit (maybe for the last three examples you mentioned). You should probably know a few of them from High School.

Now, let's try this definition on your examples.

1) $x_n = \frac{1}{n}$ (I take $k = 1$, but this is the same idea for other $k$'s...)
Let $\varepsilon > 0$. We choose $N = \left\lfloor \frac{1}{\varepsilon} \right\rfloor + 1$. Note that $N$ depends on $\varepsilon$ (not necessarily but it can and this is obvious because of the logical order of quantifiers: forall .. (fixed) there exists ((almost) one) ..). Then for all $n > N$, $|x_n - 0| = |x_n| = \frac{1}{n} < \frac{1}{N} < \varepsilon$.

Well, this is how it is written in all books.. But you stay at the same point.. Why is it so easy for him ? I know I was like you.. So one strategy is to think in the opposite way.

What $N$ should I take to have an equality $\frac{1}{N} = \varepsilon$ ? Unsurprisingly, the answer is $N = \frac{1}{\varepsilon}$. But actually, we are not sure this is an integer, so we take the integer part of this real number. But by definition, $\lfloor x \rfloor \leq x$ or equivalently, $\frac{1}{\lfloor x \rfloor} \geq 1/x$. So with the choice $N = \left \lfloor \frac{1}{\varepsilon} \right \rfloor$, we have $\frac{1}{N} \geq \varepsilon$ ! So you take $N = \left\lfloor \frac{1}{\varepsilon} \right\rfloor + 1$, to have $\frac{1}{N} < \varepsilon$, because by definition $x < \lfloor x \rfloor + 1$. Now to finish, by the type of sequence given, we see that for all $n > N$, $\frac{1}{n} < \frac{1}{N}$...

2) Let $k \geq 1$ be a real number. Then the sequence $x_n = \frac{1}{n^k}$ converges to $0$ as $n \rightarrow \infty$. So again:
Let $\varepsilon > 0$. Now we can use point 1, i.e. There exists $N \in \mathbb{N}$ such that $\frac{1}{n} < \varepsilon$ for all $n > N$. And thus, $|x_n| = \frac{1}{n^k} \leq \frac{1}{n} < \varepsilon$ for all $n > N$. So it suffices to take $N$ of point 1.

3) Finally, $x_n = \frac{4n+1}{n+3}$. Intuitively, we can write it as: $x_n = \frac{4 + \frac{1}{n}}{1 + \frac{3}{n}}$, so the limit is $4$. We can also use point 1. but let's just do it properly. Let $\varepsilon > 0$. We choose $N = \left \lfloor \frac{11}{\varepsilon} - 3 \right \rfloor + 1$ if $\varepsilon \leq \frac{11}{3}$ and $0$ otherwise. Then for all $n > N$, we have $|x_n - 4| = \left | \frac{4n+1}{n+3} - 4 \right | = \left |\frac{-11}{n+3} \right | = \frac{11}{n+3} < \frac{11}{N+3} < \varepsilon$.

Try to understand all these things.. If you have some questions, do not hesitate to ask !

For training this definition, try to prove:

If $(x_n)_{n \in \mathbb{N}}$ and $(y_n)_{n \in \mathbb{N}}$ are two sequences that converge to $x$ and $y$ respectively, then $x_n + y_n$ and $x_ny_n$ converge to $x+y$ and $xy$ respectively. (For the second one, I recall that every sequence that converges is bounded, i.e. $\exists M > 0$: $|x_n| < M$, $\forall n \in \mathbb{N}$).

Best regards,

Marcus