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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

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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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    Thanks, thats the explanation i was looking for. I am quite thick sometimes.2012-08-21
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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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    Yes but how can you just factor out n^2 like that? Wouldn't the sum of 1/(n^2) n times be = to 1/n ??2012-08-21
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    Just like the sum of (1/n) n times would equal to 1 ?2012-08-21
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    @ordinary The $\frac{1}{n^2}$ is a constant. The constant appears in each term of the finite sum, so by the distributive law, you can factor it out.2012-08-21
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    Okay. So am I wrong to assume that the sum of 1/(n^2) n times = 1/n? For example, from 1 to 5, the sum of 1/25 = 1/52012-08-21
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    @ordinary Yes $\frac{1}{n^2} n = \frac{1}{n}$. However, don't forget that you still have that $\frac{n + 1}{2}$ to multiply to it. Hence you get $\frac{n + 1}{2n}$.2012-08-21
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    So would it not be 1/n * n(n+1) / 2 == n(n+1) / 2n == (n+1)/22012-08-21
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    To fully understand what's going on, let, for example, $n=4$. Then our sum is $\frac{1}{16}+\frac{2}{16}+\frac{3}{16}+\frac{4}{16}$. This is $\frac{1}{16}(1+2+3+4)$. Now do the same thing for say $n=7$. The general case may now be clearer.2012-08-21
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    Okay, got it. Thanks fellows2012-08-21
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If you meant

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

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