Probability Theory : convergence in distribution I would need some help for a pratical exercise of probability about convergence of random variables.
Consider the following distribution function : 
$F^{X_{n}}(x) = \frac{e^{nx}}{e^{nx}+1} ;  n \geq 1$.
Proof there is  a sequence of random variables $(X_{n}) ; n \geq 1$ , which law is given for all $n$ by $F^{X_{n}}$.
Does this sequence converge in distribution ?
I cannot find such a  sequence of  random variables. Any ideas ? 
Thanks in advance for helping
 A: Since the function $F_1:x\mapsto\mathrm e^{x}/(1+\mathrm e^{x})$ is continuous and increasing from its limit $0$ at $-\infty$ to its limit $1$ at $+\infty$, it is a proper distribution function. Choose any random variable $X_1$ with distribution function $F_1$.
For every positive integer $n$, $F_n(x)=F_1(nx)$ for every real number $x$ hence $F_n$ is the distribution function of $X_n=X_1/n$. Since $X_1$ is almost surely finite, $X_n\to0$ almost surely, hence in distribution. (Alternatively, one can note that $F_n(x)\to0$ for every $x\lt0$ and $F_n(x)\to1$ for every $x\gt0$.) 
Finally, one can realize $X_1$ as $X_1=\log(U/(1-U))$, where the random variable $U$ is uniformly distributed on $(0,1)$.
A: $$F^{X_{n}}(x) = \frac{e^{nx}}{e^{nx}+1}=\int_{-\infty}^xf(t)dt$$ differentiate to get
$$
f(x)=\frac{ne^{nx}(e^{nx}+1)-ne^{nx}e^{nx}}{(e^{nx}+1)^2}=\frac{ne^{nx}}{(1+e^{nx})^2}
$$
we can check that this is a probability density
$$\int_{-\infty}^{\infty}f(x)dx=\int_1^{\infty}\frac{1}{u^2}du=1$$
