Cauchy sequences and completness of formal power series Definitions:

*

*$K[[t]] =\{\sum_{k=0}^{\infty}a_kt^k: a_k \in K\}$.

Every $a\in K[[t]] $ Can be written as $a=t^ma'$ where $a'\in K[[t]]$ is invertible ($m\geq 0$).
($a'$ is invertible iff the constant term in its expansion $\neq 0$).
Valuation on $K[[t]]$:
$v(a)$ = the minimal index $k$ such that $a_k\neq 0$.


*$K((t))=\{\frac{a}{b} : a, b\in K[[t]], b\neq 0\}$.
Element in $K((t)) $ Can be written as:
$\frac{a}{b} = t^{m-n}\frac{a'}{b'}$ ($m, n \geq 0$).

Valuation on $K((t))$:
$v(\frac{a}{b})=v(a)-v(b)$.


*Laurent  series:
$K((t))_{lor}=\sum_{k=l}^{\infty} a_kt^k$. ($l\in Z$).


*Puiseux series:
$K((t))_{pui}=\cup_{n\in N}K((t^{\frac{1}{n}}) )$ where
$K((t^{\frac{1}{n}}) =\{a(t^{\frac{1} {n}} ): a(t)\in K((t)) _{lor}\}$.
We can see that $K((t)) =K((t))_{lor}$
Question 2+3
Here are my trials for parts a, b and c:
A. $p_1(t), p_2(t),..$ is a cauchy sequence in $K[[t]]$. So for all $k\geq 0$ there is $N_k\geq 1$ such that for all $m\geq N_k$:
$|p_m-p_{N_k}|<e^{-k}$. Write:
$p_m(t)=\sum_{n=0}^{\infty} a_{m,n}t^n$.p for $m\geq 1$.
Then this is equivalent to say that:
$v(p_m-p_{N_k})>k$, so the coeficcients of $1,t,...,t^k$ are the same in $p_m(t), p_{N_k}(t)$.
Therefore $a_{m,n}=a_{N_k}$ for all $m\geq 
N_k$ and $0\leq n\leq k$. Now we assume that
$N_0<N_1<...$ and define $a_k=a_{N_k,k}$.
So,
$p_{N_k}(t)=\sum_{n=0}^{\infty}a_{N_k,n}^n=a_0+a_1t+...+a_kt^k+\sum_{n=k+1}^{\infty} a_{N_k,n}t^n$.
Finally, take $p(t)=\sum_{n=0}^{\infty}a_kt^k$
So $|p-p_m|<e^{-k}$ $\forall m\geq N_k$ thus $p_m\to p\in K[[t]]$.
B. Using similar approach as in a:
$p_1(t), p_2(t),..$ is a cauchy sequence in $K((t))$. So for all $k\geq 0$ there is $N_k\geq 1$ such that for all $m\geq N_k$:
$|p_m-p_{N_k}|<e^{-k}$. Write:
$p_m(t)=\sum_{n=l}^{\infty} a_{m,n}t^n$ where $l$ is a lower bound of the indices of all the entries $a_{m,n}$ of $p_m(t)$ ($m\geq 1$)
Then this is equivalent to say:
$v(p_m-p_{N_k})>k$, so the coeficcients of $t^l,...,t^k$ are the same in $p_m(t), p_{N_k}(t)$.
Therefore $a_{m,n}=a_{N_k}$ for all $m\geq 
N_k$ and $l\leq n\leq k$. Now we assume that
$N_l<N_{l+1}<...$ and define $a_k=a_{N_k,k}$.
So,
$p_{N_k}(t)=\sum_{n=l}^{\infty}a_{N_k,n}^n=a_lt^l+a_{l+1}t^{l+1}+...+a_kt^k+\sum_{n=k+1}^{\infty} a_{N_k,n}t^n$.
Finally, take $p(t)=\sum_{k=l}^{\infty}a_kt^k$
So $|p-p_m|<e^{-k}$ $\forall m\geq N_k$ thus $p_m\to p\in K((t))$.
Here, I have some doubt about the integer $l$ I defined. What do you think about it?
C.by contradiction, let $k>0$ and take $p(t)=\sum_{n=-l}^{\infty}t^{n/2}\in K((t))_{pui}$ then for all sequences $q_m(t)=\sum_{n=l}^{\infty} a_{m,n} t^n \in K((t))=K((t))_{lor}$ ($m\geq 1$) ($l\in Z$) how to exactly define $l$?
we have
$|p-q^m|<e^{-k}$
So $v(p-q^m)>k$.
Then, $v(p)=-l/2$ and $v(q_m) \geq l$. $v(p) \neq v(q_m)$ therefore
$v(p-q_m)=min\{v(p), v(q_m)\}$ which is $-l/2$ if $l$ is positive and $\geq l$ if $l$ is negative so $v(p-q_m)$ is not > 0 thus not > k.
I am not sure if my proof in c is enough accurate and again if $l$ is fine!.
I assume d should be seen in similar way as above,  but I am not totally sure if I did the other parts in a correct way/right thinking or not, I would prefer first to hear your opinion and advice for what I showed. This would add to my understanding.
 A: Exercise (a)
There is a typo in the statement of exercise (a). "... for every $k$, the $n_k$-the element of the sequence $p_{n_k}$ equals $a_0+a_t+\cdots+a_kt^k$" should have been "... for every $k$, the $n_k$-the element of the sequence $p_{n_k}$ equals $a_0+a_t+\cdots+a_kt^k+(\text{terms with powers greater than }k)$". In fact, you proved the correct version of that statement.
Exercise (b)
As you suspected, the introduction of  the integer $l$ as "a lower bound of the indices of all the entries $a_{m,n}$ of $p_m(t)$ ($m\geq 1$)" should be accompanied with an explanation why there is such a common lower bound for all $p_m(t), m\ge1$.
For example, if $p_m(t)=t^{-m}$ for all $m\ge1$, there is no such $l$.
Well, in fact, you have proved there is such a common lower bound implicitly.
"For all $k\geq 0$ there is $N_k\geq 1$ such that for all $m\geq N_k$: $|p_m-p_{N_k}|<e^{-k}$."
Take $k=0$. For all $m\geq N_0$, we have $|p_m-p_{N_0}|<e^0$. This is equivalent to say $v(p_m-p_{N_0})>0$. So the coefficient of $t^n$ are the same in $p_m(t), p_{N_0}(t)$ for all $n\le0$. Since the coefficient of $t^{n}$ of $p_{N_0}(t)$ is $0$ for all $n<v(p_{N_0})$, so is the coefficient of $t^n$ of $p_m(t)$ for all $n< l'$, where $l'=\min(0, v(p_{N_0})$.
Let $l=\min(l', v(p_1), v(p_2), \cdots, v(p_{N_0-1})$, which is the lower bound we wanted.
Exercise (c)
Here is an easy way. Take $p(t)=t^{-\frac12}\in K((t))_{pui}$. Let $q(t)=\sum_{n=l}^{\infty} a_n t^n$, $a_i\in K$ for all $i\ge l$, $a_l\not=0$ be an arbitrary element in $K((t))$. Consider $q(t)-p(t)=-t^{-\frac12}+\sum_{n=l}^{\infty} a_n t^n$. There are two cases.

*

*$l\le -1$. Then $v(q(t)-p(t))=l$, $|q(t))-p(t)|=e^{-l}>1$

*$l\ge 0$. Then $v(q(t)-p(t))=-\frac12$, $|q(t))-p(t)|=e^{\frac12}>1$
In all cases, $q(t)$ is outside the ball centered at $p(t)$ with radius $1$. Hence $K((t))$ is not dense in $K((t))_{pui}$.
Here is another way. We can verify that the metric of $K((t))$ is the same as the metric of $K((t))_{pui}$ restricted to $K((t))$. Since $K((t))$ is a complete metric space as proved in exercise (b), its closure in any enclosing metric space is itself. In particular, the closure of $K((t))$ in $K((t))_{pui}$ is itself. Since $K((t))\not=K((t))_{pui}$, $K((t))$ is not dense in $K((t))_{pui}$.  I do not know if this way of slightly-advanced reasoning depends on some knowledge not studied in your class.
Exercise (d)
Consider the sequence $p_i(t)\in K((t))_{pui}$, $i\in\Bbb N$, where $p_i(t)=\sum_{n=1}^{i}t^{n+\frac1n}\in K((t^{\frac1{i!}}))$. It is clear that $p_i(t)$ is a Cauchy sequence.
Intuitively, $p_i(t)$ converges to $\sum_{n=1}^{\infty}t^{n+\frac1n}$, which is not an element in $K((t^{\frac1m}))$ for any $m\in\Bbb N$.  However, this is not a proof, since the meaning of the infinite sum $\sum_{n=1}^{\infty}t^{n+\frac1n}$ is not defined yet.
Towards a contradiction, assume $p_1, p_2, \cdots$ converges to an element $q(t)\in K((t^{\frac1d}))$, where $d\in \Bbb N$.
For all $i>d$, $q(t)\in K((t^{\frac1{i!}}))$. So, we can consider $q(t)-p_i(t)$ in $K((t^{\frac1{i!}}))$. Every term in $q(t)$ has the form $$a_m(t^{\frac1d})^m=a_mt^{\frac md}$$ for some $a_m\in K$ and $m\in\Bbb Z$, i.e. the exponent of the power of $t$, $\frac md$ is a (possibly negative) multiple of $\frac1d$. Since $d+1+\frac1{d+1}$ is not a multiple of $\frac1{d}$, the coefficient of $t^{d+1+\frac1{d+1}}$ in $q-p_i$ is $-1\not=0$. We have $v(q-p_i)≤d+1+\frac1{d+1}<d+2$.
That means, for all $i>d$, $p_i$ is outside of the ball centered at $q$ with radius $e^{-(d+2)}$. Hence, the sequence $p_1, p_2, \cdots$ does not converge to $q$.
