Given a fixed $n$, we define two probabilities $p_1=\displaystyle \frac{1}{n}$ and $p_2=1-p_1 = \displaystyle \frac{n-1}{n}$. The goal is to evaluate/approximate $\displaystyle \sum_{i=1}^{n} p_1^i p_2^{n-i} $:
$$ \displaystyle \sum_{i=1}^{n} \left(\displaystyle \frac{1}{n}\right)^i \displaystyle \left(\frac{n-1}{n}\right)^{n-i}$$
Any sugesstion is much appreciated.
Hint:
Observe that $n^in^{n-i}=n^n$ is independent of $i$. This renders the summation a geometric summation.
$$\sum_{i=1}^n p_1^i p_2^{n-i} = p_2^n \sum_{i=1}^n \left ( \frac{p_i}{p_2} \right )^i = p_2^n \frac{p_1}{p_2} \frac{\displaystyle 1-\left (\frac{p_1}{p_2}\right )^n}{\displaystyle 1-\frac{p_1}{p_2} }= p_1 \frac{p_2^n-p_1^n}{p_2-p_1}$$
Plugging in your values $p_1=1/n$ and $p_2=1-1/n$, the sum is then
$$\frac1{n^{n}} \frac{(n-1)^n-1}{n-2}$$
When $n=2$, one may simply use L'Hopital or directly sum.
