How to evaluate a sum which contains limit variables? 
How to evaluate a sum which contains limit variables?

For example: $$\lim_{n\to\infty}\sum_{i=1}^n\frac{n-1}n\frac{1+i(n-1)}n $$
And would the result necessarily be rational, because each term appears to be the multiplication of two rational fractions? 

Notice that in the sum, each piece has a common factor of $\frac{n-1}{n^2}$ which can be pulled out of the sum as per the distributive property since it does not depend on the index, $i$.
$$\lim\limits_{n\to\infty}\sum\limits_{i=1}^n \frac{n-1}{n}\frac{1+i(n-1)}{n} = \lim\limits_{n\to\infty}\left(\frac{n-1}{n^2}\left(\sum\limits_{i=1}^n 1+i(n-1)\right)\right)$$
Now, what is remains in the sum can be split into two seperate sums with the one on the right again having a common factor of $(n-1)$ present which can be brought outside.
$$=\lim\limits_{n\to\infty}\left(\frac{n-1}{n^2}\left(\sum\limits_{i=1}^n 1\right)+\frac{(n-1)^2}{n^2}\left(\sum\limits_{i=1}^ni\right)\right)$$
The left sum is equal to $n$ since there are $n$ occurences of adding $1$ together, and the right sum is the $n$th triangle number, given by $\frac{n(n+1)}{2}$
$$=\lim\limits_{n\to\infty}\left(\frac{n-1}{n^2}(n)+\frac{(n-1)^2}{n^2}\left(\frac{n(n+1)}{2}\right)\right)$$
Hopefully you can take it from here.
