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 $\mathbf{(n+1)/2n}$.
 A: Your answer would be correct if you were given 
$$\left(\sum_{i=1}^n \frac{1}{n^2} \right)\left(\sum_{i=1}^n i\right).$$
The problem is that 
$$\sum_{i=1}^n \frac{i}{n^2} \not = \left(\sum_{i=1}^n \frac{1}{n^2} \right)\left(\sum_{i=1}^n i\right).$$
Instead, you should have
 $$\sum_{i=1}^n \frac{i}{n^2}  = \left( \frac{1}{n^2} \right)\left(\sum_{i=1}^n i\right),$$ 
which explains the book's answer.  This is just the distributive property of multiplication over addition.
A: It is well known that $\sum_{i = 1}^n i = \frac{n(n + 1)}{2}$. Therefore by factoring out $n^2$.
$\sum_{i = 1}^n \frac{i}{n^2} = \frac{1}{n^2}\sum_{i = 1}^n i = \frac{1}{n^2}\frac{n(n + 1)}{2} = \frac{n + 1}{2n} = \frac{1}{2} + \frac{1}{2n}$. 
A: If you meant 
$$\sum_{i=1}^n\frac{i}{n^2}=\frac{1}{n^2}\sum_{i=1}^n i=\frac{1}{n^2}\frac{n(n+1)}{2}=\frac{1}{2}\left(1+\frac{1}{n}\right)=$$
A: No. $n$ is a constant in the sum. So if you have
$\sum_{i=1}^n \frac{i}{n^2}$,
that is the same as $\frac{1}{n^2} \sum_{i=1}^n i$.
Since $\sum_{i=1}^n i = \frac{n(n+1)}{2}$,
the result is $\frac{n(n+1)}{2 n^2}
=\frac{n+1}{2n}$.
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