Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given positive integer $k$, evaluate

$$\lim_{n\to\infty}n\cdot\left(\sum_{i = 1}^n\left(\dfrac{i}{n}\right)^{k}\right)^{-1}$$

share|improve this question

2 Answers 2

up vote 7 down vote accepted

$$ \begin{align*} \lim_{n\to\infty}n\cdot\left(\sum_{i = 1}^n\left(\dfrac{i}{n}\right)^{k}\right)^{-1} &= \lim_{n\to\infty}\left(\frac1n \sum_{i = 1}^n\left(\dfrac{i}{n}\right)^{k}\right)^{-1}\\ \\ &=\left(\lim_{n\to\infty}\frac1n \sum_{i = 1}^n\left(\dfrac{i}{n}\right)^{k}\right)^{-1}\\ &= \left(\int_0^1 x^k dx\right)^{-1} \\ &= \left(\frac1{k+1}\right)^{-1} \\ &= k+1\\ \end{align*} $$

share|improve this answer
    
Uh I see! That's clever, recognizing the limit of sum as an integral. Thank you! –  darksky Aug 27 '12 at 7:11

The tricky part here is

$$\sum_{i=0}^n i^k = \frac{n^{k+1}}{k+1} + o(n^k)$$

Once you've got that

$$\lim_{n\to\infty} n \left( \sum_{i=0}^n \left(\frac{i}{n}\right)^k \right)^{-1} = \lim_{n\to\infty} n . n^{k} \left( \frac{n^{k+1}}{k+1} + \ldots \right)^{-1} = k+1$$

NOTE:

To show the top result I used

$$ S_n = S_{n-1} + n^k $$ and assumed that $S_n = T_n + \alpha n^{k+1}$. In that case

$$ T_n = T_{n-1} + \alpha n^{k+1} - \alpha (n-1)^{k+1} + n^k $$

And we get a reduction in the order iff the first two terms ($n^{k+1}$ and $n^k$) cancel, which will only occur if $\alpha = 1/(k+1)$.

However I'm sure there's a more elegant way to do that.

share|improve this answer

Your Answer

 
discard

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.