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I have several questions to ask:

1) Show increasing, find the upper bound if you can of
$sqrt{(n^2-1)}/n$.

$sqrt{(n^2-1)}/n= |n|sqrt{1-1/n^2}/n$ if $n$ is positive than $sqrt1$ else $-sqrt1$;

bound: $sqrt{(n^2-1)}/n le sqrt {n^2}/n = |n|/n=1$

To show if it is increasing should I do $frac{sqrt{(n+1)^2-1}}{n+1} ge frac{sqrt{(n)^2-1}}{n}$?

2)Prove the sequence $a_n = frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}$ has a limit

$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.

$a_n = (1-1/2)(1-1/4)…(1-1/2n)$

Cross-multiplication yields, for $kge1$,
$$
left(frac{2k-1}{2k}right)^2lefrac{2k-1}{2k+1}
$$
Therefore,
$$
begin{align}
prod_{k=1}^nleft(frac{2k-1}{2k}right)^2
&leprod_{k=1}^nfrac{2k-1}{2k+1}\
&=frac1{2n+1}
end{align}
$$
Thus,
$$
prod_{k=1}^inftyfrac{2k-1}{2k}=0
$$

For $nge1$,
$$
begin{align}
frac{frac{sqrt{n^2-1}}n}{frac{sqrt{(n+1)^2-1}}{n+1}}
&=sqrt{frac{(n+1)^3(n-1)}{n^3(n+2)}}\
&=sqrt{frac{n^4+2n^3-2n-1}{n^4+2n^3}}\[18pt]
&lt1
end{align}
$$
Thus, $frac{sqrt{n^2-1}}n$ is increasing. Furthermore,
$$
begin{align}
lim_{ntoinfty}frac{sqrt{n^2-1}}n
&=lim_{ntoinfty}sqrt{1-frac1{n^2}}\[6pt]
&=1
end{align}
$$

HINT

We have that

$$a_n=frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}=frac{(2n)!}{(2^nn!)^2}$$

then we can use Stirling's approximation.

Refer to the related

• What is the limit of $frac{prod Odd}{prod Even}?!$
  

$$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta $$
is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).

Laplace's method gives $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pileft(n+frac{1}{4}right)}}$.

The sequence is decreasing and bounded below by 0. That's why it has a limit.