An application of LDCT Consider the sequence of functions:
$$F_n (t)=\int_{-\infty}^t \underbrace{ \frac{\Gamma \left[ \left( n+1 \right)/2 \right] }{\sqrt{\pi n} \Gamma \left(n/2 \right)} \frac{1}{\left(1+y^2 /n \right)^{\left(n+1 \right)/2}}}_{f_n \left(y \right)} dy$$
We want to take the limit, as $n \to \infty$ and we want to be able to move the limit inside the integral. Thus we have to use the Lebesgue Dominated Convergence Theorem provided of course that $|f_{n} \left( y \right)|$ is dominated by an integrable function. 
Now my book suggests that this is the case because $$|f_n \left (y \right) | \leq 10 f_1 \left( y \right) $$ but this is not easy for me to see. Based on the above, $$f_1 \left(y \right)=\frac{1}{\pi} \frac{1}{1+y^2} $$
Does that inequality hold? I suppose that if we were to show that the sequence is nonincreasing then it would hold. But that might prove tricky for the function in question. For example for $n=2$ we get $$ \frac{1}{2\sqrt{2 }} \frac{1}{ \left (1+y^2 /2 \right)^{3/2}} $$ and we have to keep evaluating recursively the gamma function for all $n>1.$
Is there perhaps an easier way to understand why the above inequality is true? All suggestions are welcome. Thank you.
 A: 
and we have to keep evaluating recursively the gamma function

Separate the constant involving the $\Gamma$s from the function.
$$C_n = \frac{\Gamma \left(\frac{n+1}{2}\right)}{\sqrt{n}\Gamma\left(\frac{n}{2}\right)}$$
is almost independent of $n$, so for the dominated convergence theorem can be replaced by a suitable constant. We have
$$\frac{C_{n+2}}{C_n} = \sqrt{\frac{n}{n+2}}\cdot \frac{\frac{n+1}{2}}{\frac{n}{2}} = \frac{1+\frac1n}{\sqrt{1+\frac2n}} = 1 + O\left(\frac{1}{n^2}\right),$$
so $\lim\limits_{n\to\infty} C_{2n}$ and  $\lim\limits_{n\to\infty} C_{2n+1}$ exist and are nonzero.
Thus it remains to see that
$$\frac{1+y^2}{\left(1 + \frac{y^2}{n}\right)^{(n+1)/2}}$$
is bounded independent of $n$.
The mean value theorem yields
$$\left(1+\frac{y^2}{n}\right)^{(n+1)/2} \geqslant 1 + \frac{n+1}{2n}y^2 > 1 + \frac12y^2,$$ and that gives us
$$\frac{1+y^2}{\left(1 + \frac{y^2}{n}\right)^{(n+1)/2}} < 2$$
for all $n$ and $y$.
For the explicit constant $f_n(y) \leqslant 10\cdot f_1(y)$, we must estimate the quotient $\dfrac{C_n}{C_1}$, but to apply the dominated convergence theorem it suffices that the ratio is bounded, which follows from the existence of the limit(s).
