Limit of a series in probability class I  saw the following question in previous exams, and cant realy solve it..
There are n+1 buckets, from 0 to n.
In the I-th bucket there are n balls, i white and n-i black.
Picking randomly with equal chance a bucket, and pulling 3 balls from it (after we take a ball, we return it), so we pull the first ball, return it, then pull the second, return it, then the third.
An= The probability that all 3 balls are white.
$\displaystyle{\lim_{n \to \infty}} An$ =?
Answer: 0.25
I tried solving it and got to:
$\sum (\frac{i}{n})^{3} \cdot  (\frac{1}{n+1})$
but im not sure how to advance from that.
Sorry for English mistakes and thanks for the help!
 A: The first thing to note is that the summation can be written as 

$$\frac{1}{n^3(n+1)} \sum_{i=1}^{n}i^3$$

which greatly simplifies the problem to evaluating the function 

$$f(k,n) = \sum_{i = 1}^{n} i^k$$

at $k = 3$, i.e. evaluate $f(3,n)$. Luckily for us there is a nice recurrence relation that solves the problem, notice that 

$$(n+1)^{k} - n^k = \sum_{i=0}^{k-1}\binom{k}{i}n^i$$

implies that 

$$\frac{\Delta f}{\Delta y}\bigg|_{(k,m)}  =f(k,m+1)-f(k,m)=\sum_{n=0}^{n=m}(n+1)^{k} - n^{k}= \sum_{n=0}^{n=m}\sum_{i=0}^{k-1}\binom{k}{i}n^i = \sum_{i=0}^{k-1}\binom{k}{i} f(i,m)$$

which implies that 

$$\sum_{j=1}^{j=m-1}\frac{\Delta f}{\Delta y}\bigg|_{(k,j)}   = f(k,m)   = \sum_{j=1}^{j=m-1} \sum_{i=0}^{k-1}\binom{k}{i} f(i,j)$$

We now solve recursively, we know that $f(0,n) = \sum_{i=1}^n i^0= n $ which allows us to solve for $k=1$

$$f(1,n)  = \sum_{j=1}^{j=m-1} \sum_{i=0}^{0}\binom{1}{i} f(i,j)  = \sum_{j=1}^{j=m-1} f(0,j) = \sum_{j=1}^{j=m-1} j = \binom{m}{2}$$

Solving for $k=2$ we get 

$$ f(2,m)   = \sum_{j=1}^{j=m-1} \sum_{i=0}^{2-1}\binom{2}{i} f(i,j) = \sum_{j=1}^{j=m-1}  \left( \binom{2}{0} j  \binom{2}{1} \binom{j}{2}  \right) = \frac{n(n+1)(2n+1)}{6}$$

performing a similar computation for $k=3$ we get 

$$f(3,n) = \left(\frac{n(n+1)}{2}\right)^2$$

and therefore the limit is 

$$\lim_{n \to \infty} A_n = \lim_{n \to \infty} \frac{1}{n^3(n+1)} \left(\frac{n(n+1)}{2}\right)^2\lim_{n \to \infty} \frac{n+1}{4n} = \frac{1}{4}$$

