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The following expression

$\lim_{n\to\infty} \sum_{i=1}^{n} \frac{4}{n}\cdot \frac{4+4i}{n}$

can (according to the book I'm reading, and I'm sure it's correct) be simplified to

$\lim_{n\to\infty} \sum_{i=1}^{n}\frac{16(n+i)}{n^2}.$

Where is the numerator $n$ coming from? Looking at it it seems like it should simplify to

$\lim_{n\to\infty} \sum_{i=1}^{n}\frac{16(1+i)}{n^2}$

What painfully obvious fact am I ignoring?

UPDATE

In hindsight (and with the answers here) I believe it is a typo, but should in fact read

$\lim_{n\to\infty} \sum_{i=1}^{n} \frac{4}{n}\cdot \Big( 4+ \frac{4i}{n} \Big)$

Which does simplify to $\lim_{n\to\infty} \sum_{i=1}^{n}\frac{16(n+i)}{n^2}.$

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    Typo! - at least at first look. They happen more often in math books than you might think. Convergence behavior of the first term at least changes radically going from $\frac{1}{n^2}$ to $\frac{1}{n}$, so I doubt there is a clever way to justify this.2012-12-28

2 Answers 2

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You are correct, but you haven't done enough algebra to show the book is incorrect: $\begin{align}\lim_{n\to\infty}\frac{16}{n^2}\sum_{i=1}^n(1+i)&=\lim_{n\to\infty}\frac{16}{n^2}\left[\sum_{i=1}^n1+\sum_{i=1}^ni\right]\\ &=\lim_{n\to\infty}\frac{16}{n^2}\left(n+\frac{n(n+1)}{2}\right)\\ &=\lim_{n\to\infty}\left(\frac{16}{n}+8\frac{n^2+n}{n^2}\right)\\ &=8. \end{align}$Now, by the book, the limit would have to be at least $16$ since $\lim_{n\to\infty}\sum_{i=1}^n\frac{16n}{n^2}=16$ and we didn't take into account the positive $i$ terms that are added.

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    @franklin: The harmonic series is in fact $\sum_{k=1}^\infty\frac{1}{k}$. You have to check your indices to see that what I have above is correct.2012-12-29
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Hint: use the identities

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

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    @AustinMohr: So, what do you think it has been used in the answer posted by Clayton?2012-12-28