Showing that such sequence is monotone. I have trouble figuring out how to show that 
$\frac{\sqrt{n+1}}{(n+1)^{2}+1}<\frac{\sqrt{n}}{n^{2}+1},\quad\forall n\geq1$.
I have solved several inequalities before, but I can't prove this one.
Thanks!
 A: Your inequality is equivalent, after multiplying through by the product of denominators, and squaring the (positive) sides, to
$$n((n+1)^2+1)^2>(n+1)(n^2+1)^2.$$
In this if the sides are expanded one arrives at the inequality
$$3n^4+6n^3+6n^2+3n-1>0,$$ clearly true for positive $n.$
A: This is equivalent to
$\sqrt{1+1/n} < ((n+1)^2+1)/(n^2+1)$.
Since $\sqrt{1+1/n} < 1+1/(2n)$
and $((n+1)^2+1)/(n^2+1) = 1+(2n+1)/(n^2+1)$,
this is true if
$1/(2n) < (2n+1)/(n^2+1)$
or $4n^2+2n > n^2+1$,
which is true for $n \ge 1$.
Note that more than this is true: Let's try to find the largest value of $c > 1$
such that 
$\sqrt{1+c/n} < ((n+1)^2+1)/(n^2+1)$.
As before,
$\sqrt{1+c/n} < 1+c/(2n)$
so we want
$c/(2n) < (2n+1)/(n^2+1)$
or $n^2+1 < (2/c)n(2n+1)
=(4/c)n^2+2/c$.
For $c = 4$ this is false,
but it is true for large enough $n$
for any $c < 4$.
To see this, let $4/c = 1+d$ where $d > 0$.
Then we want 
$n^2+1 < (1+d)n^2 + (1+d)/2$
or $d n^2 > 1-(1+d)/2 = (1-d)/2$
or, assuming $d < 1$ (this is true for all $n$ if $d \ge 1$),
$n^2 > (1-d)/(2d) = 1/(2d) -1/2$.
Thus $n > \sqrt{1/2d}$ works.
So, there is no largest value of $c$ 
such that
$\sqrt{1+c/n} < ((n+1)^2+1)/(n^2+1)$
(at least with this elementary method),
but the inequality is true for any $c < 4$ for all large enough $n$.
A: Let 
$$ f(x)=\frac{\sqrt{x}}{x^2+1}. $$
Then it is easy to check
$$ f'(x)=\frac{-3x^2+1}{2\sqrt{x}(x^2+1)}\le 0 $$
for $x>1$. It means that $f(x)$ is strictly decreasing. so $f(n+1)< f(n)$ or 
$$ \frac{\sqrt{n+1}}{(n+1)^2+1} < \frac{\sqrt{n}}{n^2+1}. $$
