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 April 2nd, 2011, 11:33 AM #1 Newbie   Joined: Apr 2011 Posts: 10 Thanks: 0 Limit Let $k$ be a positive integer, $\alpha$ is an arbitrary real number. Find the limit of sequence $a_{n}$ where: $a_{n}=\frac{\left [1^{k}.\alpha\right ]\left [2^{k}.\alpha\right ]+..+\left [n^{k}.\alpha\right ]}{n^{k+1}}$ $(n=1,2,...),$ here the notation $\left [ x \right ]$ is the largest integer that does not exceed $x$.
 April 2nd, 2011, 04:41 PM #2 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Limit I think you made a mistake in your equation. I'm not sure what is meant by .a, and you may have forgotten a + sign in the numerator. Do you mean the limit at n approaches infinity?
 April 2nd, 2011, 04:43 PM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Limit Many use a decimal to denote multiplication.
 April 2nd, 2011, 04:58 PM #4 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Limit Assuming the numerator is actually $\sum_{j=1}^n\lfloor\alpha j^k\rfloor,$ very quick jottings (at 2 in the morning, so I assume no responsibility for the lack of accuracy in my analysis ) suggest that the limit as $n\to\infty$ is $\alpha/(k+1).$
 April 2nd, 2011, 05:53 PM #5 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Limit If that's true, then as n approaches infinity, the limit may be zero.
 April 2nd, 2011, 06:47 PM #6 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Limit I was wrong. Doing some research, I found Faulhaber's Formula which can be used to determine the formula for these types of sums in terms of n. Both the numerator and denominator are of degree k+1, therefore the limit at infinity does exist. The exact limit of this problem as n goes to infinity is floor(a)/(k+1) after determining the coefficient of the n^(k+1) term in the numerator.
 April 3rd, 2011, 07:30 AM #7 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Limit I don't think so - $\frac1{n^{k+1}}\sum_{j=1}^n(\alpha j^k-1)\,<\,\frac1{n^{k+1}}\sum_{j=1}^n\lfloor\alpha j^k\rfloor\,<\,\frac1{n^{k+1}}\sum_{j=1}^n\alpha j^k.$ Both the left hand and right hand expressions are of the form $\frac\alpha{k+1}+\mathcal{O}$$1/n$$,$ so the expression in the middle must converge to $\frac\alpha{k+1}$ rather than $\frac{\lfloor\alpha\rfloor}{k+1}.$
 April 3rd, 2011, 10:14 AM #8 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Limit Interesting use of the squeeze theorem. I hadn't thought of that.

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