Tail Probability for Student's t-Distribution Let $n$ be a positive integer, $t < 0$, and let $F_n$ the cumulative distribution function of Student's t-distribution with $n$ degrees of freedom. Then it is a well known fact (which I always found stated without proof) that the sequence $(F_n(t))_{n=1}^{\infty}$ is strictly decreasing. 
Do you know some proof of the fact that $(F_n(t))_{n=1}^{\infty}$ is a strictly decreasing sequence?
NOTE. Note that since we have $F_n(t) \rightarrow \Phi(t)$ as $n \rightarrow \infty$, where $\Phi$ is the cumulative distribution function of the standard normal distribution, from the proof of the monotonicity property above we would get the well known inequality $F_n(t) > \Phi(t)$ for all $n$ (which I found always quoted without proof, too).
 A: I finally found the proof, by following Henry's suggestion.
For every positive integer $m$, let $f_m(t)$ be the density function of Student's t-distribution with $m$ degrees of freedom, that is
\begin{equation}
f_m(t)=\frac{\Gamma \left(\frac{m+1}{2} \right)}{\sqrt{m \pi} \Gamma \left( \frac{m}{2} \right)} \left( 1 + \frac{t^2}{m} \right)^{-\frac{m+1}{2}} \quad (t \in \mathbb{R}).
\end{equation}
Fix a positive integer $n$ and define the ratio
\begin{equation}
R(t)=\frac{f_{n+1}(t)}{f_{n}(t)} \quad (t \in \mathbb{R}).
\end{equation}
First let us note that 
\begin{equation}
\lim_{t \rightarrow - \infty} R(t) =0.
\end{equation}
Moreover, we have
\begin{equation}
R'(t)=R(t) \frac{t(1-t^2)}{n(n+1) \left(1+\frac{t^2}{n} \right) \left(1+\frac{t^2}{n+1} \right)} \quad (t \in \mathbb{R}),
\end{equation}
so that $R'(t) > 0$ for $t < -1$ and $R'(t) <0$ for $-1 < t < 0$. 
Now, let us note that since $F_n(0)=F_{n+1}(0)=1/2$, there must exist at least one value $\bar{t} < 0$ such that $R(\bar{t})=1$. Clearly we must have $\bar{t} < -1$, and since $R(t)$ is strictly increasing on $(-\infty,-1]$, there exist only one $\bar{t} \in (-\infty,-1]$ such that $R(\bar{t})=1$. Now note that $R(t)$ is strictly decreasing on $(-1,0]$, so in order to prove that there is no $s \in (-1,0]$ such that $R(s)=1$, it is enough to prove that $R(0) > 1$.
In order to prove that $R(0)>1$, we shall us two remarkable inequalities proved by Wallis in his great work "Arithmetica Infinitorum": for any positive integer $k$, we have
\begin{equation}
\left[\frac{(2k-1)!!}{(2k-2)!!}\right]^2 \frac{1}{2k} \sqrt{\frac{2k+1}{2k}} < \frac{2}{\pi} < \left[\frac{(2k-1)!!}{(2k-2)!!} \right]^2 \frac{1}{2k}\sqrt{\frac{2k}{2k-1}}. \tag{I}
\end{equation}
For a short proof and very elegant proof, see the work by Dutka On Some Gamma Function Inequalities.
Now, assume that $n$ is even. Then we have
\begin{equation}
R(0)=\left(\frac{2 \cdot 4 \cdot 6 \cdot \dots \cdot (n-2)}{3 \cdot 5 \cdot \dots \cdot (n-1)} \right)^2 \frac{2 n \sqrt{n}}{\pi \sqrt{n+1}}.
\end{equation}
The inequality $R(0) > 1$ can be so written:
\begin{equation}
\frac{\pi}{2} < \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \dots \cdot \frac{n-2}{n-3} \cdot \frac{n-2}{n-1} \cdot \frac{n}{n-1} \cdot \sqrt{\frac{n}{n+1}},
\end{equation}
which is the first inequality in (I).
Analogously, if $n$ is odd, we get
\begin{equation}
R(0)=\left(\frac{3 \cdot 5 \cdot 7 \cdot \dots \cdot (n-2)}{2 \cdot 4 \cdot \dots \cdot (n-1)} \right)^2 \frac{\pi n \sqrt{n}}{2 \sqrt{n+1}}.
\end{equation}
The inequality $R(0) > 1$ now reads
\begin{equation}
\frac{\pi}{2} > \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \dots \cdot \frac{n-1}{n-2} \cdot \frac{n-1}{n} \cdot \frac{n+1}{n} \cdot \sqrt{\frac{n}{n+1}},
\end{equation}
which is the second inequality in (I).
So, we have that $R(t) < 1$ for $t < \bar{t}$ and $R(t) > 1$ for $\bar{t} < t < 0$. This result, when combined to the fact that $F_n(0)=F_{n+1}(0)=1/2$, allows us to conclude that $F_{n+1}(t) < F_{n}(t)$ for each $t < 0$.
QED
