An urn contains balls numbered 1 to N. Let X be the largest number drawn in n drawings when random sampling with replacement is used. (The event X k means that each of n numbers drawn is less than or equal to k.) Show that when N is large:
$${E[X] {\approx} \frac{n * N}{n+1}}$$
Here is my approach:
$$P(X=k) = P(X \leq k) - P(X \leq k-1) $$
since $$P(X \leq k) = \frac{k^n}{N^n}$$ and $$P(X \leq k-1) = \frac{(k-1)^n}{N^n}$$ then $$E(X = k) = \sum_{k=1}^N\frac{k*k^n}{N^n} - \sum_{k=1}^N\frac{k*(k-1)^n}{N^n}$$
By using [telescoping sum][1]telescoping sum and [Riemann integral][2]Riemann integral, I am getting the answer:
$${E[X] {\approx} \frac{N}{n+1}}$$
where I don't have (n) term on the numerator.
What is wrong in my approach? Thanks.
Disclaimer: This is part of my "Probability" HW. [1]: http://mathworld.wolfram.com/TelescopingSum.html [2]: https://en.wikipedia.org/wiki/Riemann_integral
