Find $\lim\limits_{n\to \infty}{{n} \choose {k}}\left(\frac{\lambda}{n}\right)^k\left(1 - \frac{\lambda}{n}\right)^{n-k}$

Fix positive $\lambda$ so that $\frac{\lambda}{n} \in [0,1]$ for all $n \ge 1$. Next, let $P_n(k)$ denote the probability of recording exactly $k \le n$ heads in an $n$-toss experiment, where we assume that the probability of obtaining heads in any single toss is $\frac{\lambda}{n}$. An explicit formula for $P_k(n)$ is given by

$$P_n(k) = \displaystyle {{n} \choose {k}}\Big(\frac{\lambda}{n}\Big)^k\Big(1 - \frac{\lambda}{n}\Big)^{n-k}.$$

Show that $\displaystyle \lim_{n \to \infty} P_n(k)$ exists and determine its value.

I expect that $P_n(k) \to 0$, independent of what $k$ is. This is because, as the number of tosses in the experiment approaches infinity, the probability of obtaining any one particular outcome should approach zero. I have tried to adapt the solutions to the following problem: Determine $\displaystyle \lim_{n \to \infty}{{n} \choose {\frac{n}{2}}}\frac{1}{2^n}$, where each $n$ is even. Is it also possible to apply Stirling's Approximation to this limit?

Hints or solutions are greatly appreciated.

-
Did's answer here may help math.stackexchange.com/questions/174557/… –  Byron Schmuland Jan 31 '13 at 0:04
It will be $e^{-\lambda}\frac{\lambda^k}{k!}$, not $0$. –  André Nicolas Jan 31 '13 at 0:04

Note that $$\left(1 - \dfrac{\lambda}n\right)^{n-k} \sim \dfrac{e^{-\lambda}}{\left(1 - \dfrac{\lambda}n \right)^k}$$ Hence, $$\dbinom{n}k \left(\dfrac{\lambda}n \right)^k \left(1 - \dfrac{\lambda}n\right)^{n-k} \sim \dbinom{n}k \left(\dfrac{\lambda}{n - \lambda} \right)^k e^{-\lambda} = \dfrac{\lambda^k}{k!} e^{-\lambda} \dfrac{n(n-1)(n-2) \cdots(n-k+1)}{(n- \lambda)^k}$$ Now note that $$\dfrac{n(n-1)(n-2) \cdots(n-k+1)}{(n- \lambda)^k}$$ is nothing but $$\dfrac{1 \times \left(1 - \dfrac1n \right) \times \left(1 - \dfrac2n \right) \times \left(1 - \dfrac3n \right) \times \cdots \times \left(1 - \dfrac{k-1}n \right)}{\left(1 - \dfrac{\lambda}n \right)^k} \sim 1$$ Hence, your expression tends to $\dfrac{\lambda^k}{k!} e^{-\lambda}$.
Here another way to proof $\lim\limits_{n\to \infty}{{n} \choose {k}}\left(\frac{\lambda}{n}\right)^k\left(1 - \frac{\lambda}{n}\right)^{n-k}= e^{-\lambda}\dfrac{\lambda^k}{k!}$
Note that $$\frac{\lambda ^k}{k!}\left(1- \frac{\lambda}{n}\right)^{n-k} \geq {{n} \choose {k}}\left(\frac{\lambda}{n}\right)^k\left(1 - \frac{\lambda}{n}\right)^{n-k} \geq \frac{\lambda ^k}{k!}\left(1- \frac{k}{n}\right)^k\left(1- \frac{\lambda}{n}\right)^{n-k}$$ As $n \to \infty$ both R.H.S and L.H.S equal to $e^{-\lambda}\dfrac{\lambda^k}{k!}$.