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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 $\mathbf{(n+1)/2n}$.

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    The $i$ is varying in the sum. Where did it go between lines 2 and 3?2012-08-21

4 Answers 4

5

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.

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    no you are not thick2012-08-21
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It is well known that $\sum_{i = 1}^n i = \frac{n(n + 1)}{2}$. Therefore by factoring out $n^2$.

$\sum_{i = 1}^n \frac{i}{n^2} = \frac{1}{n^2}\sum_{i = 1}^n i = \frac{1}{n^2}\frac{n(n + 1)}{2} = \frac{n + 1}{2n} = \frac{1}{2} + \frac{1}{2n}$.

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    Okay, got it. Thanks fellows2012-08-21
1

If you meant

$\sum_{i=1}^n\frac{i}{n^2}=\frac{1}{n^2}\sum_{i=1}^n i=\frac{1}{n^2}\frac{n(n+1)}{2}=\frac{1}{2}\left(1+\frac{1}{n}\right)=$

1

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 $\frac{n(n+1)}{2 n^2} =\frac{n+1}{2n}$. $