We have $f_n(x) = c_n(1-\frac{x}{n})^n \mathbb{1}_{[0,n]}$ and $x \in \mathbb{R}, n\geq 1$. I want to find $c_n$ so that $\int f_n\, dm = 1$ where $m$ is Lebesgue measure and show that $\lim_{n\to\infty} f_n(x) = f(x)$ exists for all $x \in \mathbb{R}$ and that $f$ is a probability density function (pdf).
So $$1 = \int f_n \, dm = \int c_n(1-\frac{x}{n})^n \mathbb{1}_{[0,n]}, \, dm = c_n(1-\frac{x}{n})^n *n $$
So $c_n =\frac{1}{n(1-\frac{x}{n})^n}$. But then $$ \lim_{n\to\infty} f_n(x) = \lim_{n\to\infty} \frac{1}{n(1-\frac{x}{n})^n} *(1-\frac{x}{n})^n \mathbb{1}_{[0,n]} = 0 = f?$$
So how could $f$ be a probability density function if it is 0? I believe that my $c_n$ is wrong and I don't really know where to make use of $\lim_{n\to\infty} (1-\frac{x}{n})^n =e^{-x}$ or if I am calculating the integral correctly. I'll appreciate any pointers for how to go about the problem.
You need to calculate $\int_{\mathbb{R}} (1-\frac{x}{n})^n 1_{[0,n]}dx=\int_0^n (1-\frac{x}{n})^n dx=n\int_0^1 u^n du=\frac{n}{n+1}$. In your calculation, you never computed the integral correctly.
As far as what does $f_n$ converge to: Look up the definition of $e^x$ with limits.