Evaluate sum of the binomial-type distribution

Question

I have the following binomial-type distribution:

$$f(k;n,p) = \frac{1}{2^k}\binom{n}{k}p^k(1-p)^{n-k},$$

such that as $n \to \infty$ and $p \to 0$ the product $np = \lambda$ is constant. I would like to evaluate the following sums in the limit:

$$\lim\limits_{n\to\infty}\sum_{k=1}^{n}f(k;n,p),$$

where $p$ is fixed, and

$$\lim\limits_{n\to\infty, p \to 0}\sum_{k=1}^{n}f(k;n,p).$$

Answer 1

$\sum_{k=1}^nf(k;n,p)=\left(1-\frac{p}{2}\right)^n-(1-p)^n=\left(1-\frac{\lambda}{2n}\right)^n-\left(1-\frac{\lambda}{n}\right)^n$ As $n\rightarrow\infty$, this goes to $e^{\frac{-\lambda}{2}}-e^{-\lambda}$.