Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


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

share|cite|improve this question
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 3:05
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 3:05
up vote 4 down vote accepted

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.

share|cite|improve this answer
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 5:03
@franklin: I'm not sure I understand what you're saying/asking..? – Clayton Dec 28 '12 at 6:00
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 5:03

Hint: use the identities

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

share|cite|improve this answer
The term of $i$ is still present in the book's answer, so this identity has not been invoked. – Austin Mohr Dec 28 '12 at 3:04
@AustinMohr: So, what do you think it has been used in the answer posted by Clayton? – Mhenni Benghorbal Dec 28 '12 at 3:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.