Limit class of ugly sequence with integral $\renewcommand{\d}[1]{\operatorname{d}\!{#1}}$
I am to take Analysis 4 in a few days and running through old exam exercises I found this gem:

Determine the limit class of the following sequence:
$$\left \{ (-n)^{n+1} \int_0^{+\infty} \frac{1-\cos \sqrt{x/n}}{(n+x^2)^n}\d x \right \}_{n=1}^\infty $$
  (It may be useful to know that, for any fixed t, the sequence $\{(1+t/n)^n\}_{n=1}^\infty$ is increasing.

And here is what I've been able to do:
First of all, the sequence is clearly alternating, since everything in there is positive except for that $(-n)^{n+1}$ term. So I just looked at the positive half, expecting the negative to do the same, but mirrored around $0$.
Then I split the generic term of the sequence in two:
(let $f$ be the argument of the integral)
$$ (-n)^{n+1} \int_0^{+\infty} f(x)\d x =\\ 
= (-n)^{n+1} \left( 
\int_0^1 f(x)\d x + 
\int_1^{+\infty} f(x)\d x
\right)$$
Now, the first part is easily controlled:
For $z \to 0$,
$$ \cos(z)\geq 1-\frac12z^2 $$
and therefore, 
$$ \int_0^1 f(x)\d x \leq \int_0^1 \frac{\frac12 \frac xn}{n^n (1+x^2/n)^n} \d x $$
And since, as per the hint, $(1+x^2/n)^n$ is increasing,
$$n^{n+1} \int_0^1 \frac{1-\cos \sqrt{x/n}}{(n+x^2)^n}\d x \leq \\
\leq n^{n+1} \int_0^1 \frac{x}{2n^{n+1}(1+x^2/n)^n}\d x = \\
= \frac12 \int_0^1 \frac{x}{(1+x^2/n)^n}\d x \leq \\
\leq \frac12 \int_0^1 \frac x{1+x^2} \d x = \frac 14 \log(2)$$
I also thought I had a good lower bound, but in typing I found out it was nonsense.
As for the second half, I have been extremely successful at showing that its absolute value is greater than zero and smaller than infinity.
Since I do not give up easily, I have written a quick-and-dirty matlab script to try and get a feel at what this damn thing does, and here's the result:

... Well, at least it is symmetrical.
It also appears to go to to $1/2$, which means that the second part should oscillate by about $0.2$...
Anybody willing to help? Please?

The answer will turn out to be the limit class $\{-\frac14, \frac14\}$.  Here is how to show that if 
$$ a_n = (-n)^{n+1} \int_0^{+\infty} \frac{1-\cos \sqrt{x/n}}{(n+x^2)^n}\d x  
$$
then for all $n>1$, $|a_n| \leq \frac14 \frac{n}{n-1}$:
As you pointed out, for $x>0$,
$$ \cos\left(\sqrt{\frac{x}{n}} \right) \geq 1-\frac{x}{2n} \implies 
1-\cos\left(\sqrt{\frac{x}{n}} \right) \leq \frac{x}{2n}
$$
So
$$|a_n| \leq n^n\int_0^\infty n \frac{x}{2n} (n+x^2)^{-n} dx = 
n^n\left[ \frac{(n+x^2)^{-n+1}}{-n+1} \frac{1}{2n}\frac12 \right]_0^\infty
$$
The $x=\infty$ end is zero (since $n>1$ the $(x^2)^{-n+1}$ becomes smaller than $\frac{1}{n^n}$ for sufficiently large $x$).  The zero end then gives
$$
|a_n| \leq -n^n\frac14 n^{-n}\frac{n}{-n+1} =   \frac14 \frac{n}{n-1}
$$
In a similar way, you can show that $|a_n| \geq \frac14$ and that implies that the limit will be $\frac14$.
