Evaluate $\sum_{i=1}^n \frac{i}{n^2}$ My book asks to simplify this problem:
$$\sum_{i=1}^n\frac{i}{n^2}$$
This equals:  
$$\sum_{i=1}^n\frac{1}{n^2}i$$
Now, shouldn't 
$$\sum_{i=1}^n\frac{1}{n^2}=\frac{1}{n}\text{ ?}$$
Because you are summing $1/n^2$, $n$ times, so that $n / n^2 = 1/n$.
So it should be 
$$\frac{1}{n}\sum_1^n i = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \mathbf{\frac{n+1}{2}}$$
But the answer in my book is $\mathbf{(n+1)/2n}$.

It is well known that $\sum_{i = 1}^n i = \frac{n(n + 1)}{2}$. Therefore by factoring out $n^2$.
$\sum_{i = 1}^n \frac{i}{n^2} = \frac{1}{n^2}\sum_{i = 1}^n i = \frac{1}{n^2}\frac{n(n + 1)}{2} = \frac{n + 1}{2n} = \frac{1}{2} + \frac{1}{2n}$. 
