Use the comparsion criterion to show that: $$\lim_{n\rightarrow\infty}{\sqrt[n]{1+\pi^n}}=\pi \quad \text{and} \quad \lim_{n\rightarrow\infty}{(1+(1/n))^n}=e$$
I know the classical proof of this facts, but i dont know how to prove with the comparsion criterion, I've already find a lower bound for the first but i need the upper bound and the bounds for the second. Any hint? Please...
The second one is just one of the usual definitions of the number $\;e\;$ : you just prove the left hand is a convergent sequence.
For the first one use the squeeze theorem:
$$\pi=(\pi^n)^{1/n}\le\left(1+\pi^n\right)^{1/n}\le(2\pi^n)^{1/n}=\sqrt[n]2\,\pi\xrightarrow[n\to\infty]{}1\cdot\pi=\pi$$
For the second one, we proceed as follows.
In THIS ANSWER, I showed using Bernoulli's Inequality that the sequence $\left(1+\frac1n\right)^n$ is monotonically increasing. To show that it is bounded, we have using the Binomial Theorem
$$\begin{align} \left(1+\frac1n\right)^n&=1+1+\frac1{2!}\left(1-\frac1n\right)+\cdots +\frac1{n!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots \left(1-\frac{n-1}n\right)\\\\ &\le 2+\frac12\sum_{k=0}^\infty \frac{1}{2^{n}}\\\\ &=3 \end{align}$$
Thus, we have established that the sequence $\left(1+\frac1n\right)^n$ converges to a number that is less than $3$ and greater than $2$.
As for the first $\pi=\sqrt[n]{\pi^n}\leq\sqrt[n]{1+\pi^n}\le\sqrt[n]{2\pi^n}=\sqrt[n]2\sqrt[n]{\pi^n}=\sqrt[n]2.\pi$
$\lim_{n\to\infty}\pi=\pi$
$\lim_{n\to\infty}(\sqrt[n]2.\pi)=\lim_{n\to\infty}(\sqrt[n]2).\pi=1\pi=\pi$
So by the comparison test $\pi\leq\lim_{n\to\infty}\sqrt[n]{1+\pi^n}\leq\pi$
Whence $\lim_{n\to\infty}\sqrt[n]{1+\pi^n}=\pi$
For the second it really all depends on how you define $e$. You will need to say.
A common (and I think the first) definition is
$e=\lim_{n\to\infty}(1+\frac{1}{n})^n$
in which case it all becomes rather easy:
$(1+\frac{1}{n})^n\leq(1+\frac{1}{n})^n\leq(1+\frac{1}{n})^n$
So by the comparison test.
$e\leq\lim_{n\to\infty}(1+\frac{1}{n})^n\leq e$, hence etc.
P.S. I've never heard of a "comparison criterion" for sequences. There is a well known "comparison test" for series. I've assumed it meant what another answer refers to as the "squeeze theorem".
Since other answers have adequately dealt the first sequence I propose to deal with $a_{n} =(1+(1/n))^{n}$ here. The question asks us to show that $a_{n} \to e$ which suggests that there is some other definition of $e$ at work here. I chose one definition $$e=1+\frac{1}{1!}+\frac{1}{2!}+\dots\tag{1}$$ and let $$b_{n} = 1+\frac{1}{1!}+\frac{1}{2!}+\dots +\frac{1}{n!}\tag{2}$$ Clearly $b_{n} \to e$ by definitions $(1)$ and $(2)$.
Consider another sequence $$c_{n} =\left(1-\frac{1}{n}\right)^{-n}\tag{3}$$ Using binomial theorem for positive integer index we can show that $a_{n} \leq b_{n} $ and using binomial theorem for general index we can show that $b_{n} \leq c_{n} $ and therefore we can write $$a_{n} \leq b_{n} \leq c_{n} \tag{4}$$ Also note that using binomial theorem for positive integer index we can show that $a_{n} $ is increasing and clearly $b_{n} $ is also increasing and $b_{n} \to e$ hence it follows that $a_{n} \leq e$. It follows that $a_{n} $ is convergent.
Similarly via binomial theorem for general index we can show that $c_{n} $ is decreasing and obviously positive so that $c_{n} $ is convergent. Now using Bernoulli's inequality we have $$1-\frac{1}{n}\leq\left(1-\frac{1}{n^{2}}\right)^{n}=\frac{a_{n}}{c_{n}}\leq 1\tag{5}$$ and using Squeeze theorem on the above equation we can see that $a_{n} /c_{n} \to 1$. It follows that sequences $a_{n}, c_{n} $ tend to same limit and by applying Squeeze theorem on equation $(4)$ we see that all the sequences $a_{n}, b_{n}, c_{n} $ tend to the same limit which is $e$ and our job is done.
Note: I assume that comparison criterion is the same thing as Squeeze theorem.
HINT
The first is a strictly decreasing sequence in $n$, the second -- strictly increasing.
