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

Alex
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  • The simplification the book provides (as you have written it) is incorrect. Are you sure you've copied it correctly to your question? – Austin Mohr Dec 28 '12 at 03:05
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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. – gnometorule Dec 28 '12 at 03:05

2 Answers2

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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.

Clayton
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  • A few things. Firstly, I believe that the book is in error. Secondly, it seems that @Clayton the last line used which shows $\lim$ – franklin Dec 28 '12 at 05:03
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    @franklin: I'm not sure I understand what you're saying/asking..? – Clayton Dec 28 '12 at 06:00
  • haha sorry. I hit send by accident before I was done commenting. I think the last line which shows: $\lim_{n \to \infty} \sum_{i = 1} ^n 16n/(n^2) $ is a little bit ambiguous. From elementary calculus it should be known that the harmonic series is divergent i.e. $\lim_{n \to \infty} \sum (1/n) = \infty $ the last line should simplify to $\lim_{n \to \infty} 16 \sum_{i = 1} ^n 1/n \to \infty$ i.e. its divergent. not convergent to 16 as this post suggests? – franklin Dec 28 '12 at 22:45
  • @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. – Clayton Dec 29 '12 at 05:03
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Hint: use the identities

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

Mhenni Benghorbal
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