CPL.Luke Posted July 21, 2006 Share Posted July 21, 2006 this is a wierd limit, its the first time I've ever seen a limit of a function involving a function of a series I think it diverges to infinity, but I don't know how to get that mathmatically, so if someone could tell me how to take the limit of a function such as this it would be greatly appreciated. heres the function [MATH]\lim_{n\rightarrow \infty}\frac{n}{\sqrt{\sum_{n=1}^{\infty} n^2}}[/MATH] Link to comment Share on other sites More sharing options...

Invader_Gir Posted July 21, 2006 Share Posted July 21, 2006 I *think* that it would be zero, since it would be a larger number divided by an even larger number....however, I am no math genius. Link to comment Share on other sites More sharing options...

NeonBlack Posted July 21, 2006 Share Posted July 21, 2006 [math] \lim_{n\rightarrow \infty}\frac{n}{\sqrt{\sum_{n=1}^{\infty} n^2}} [/math] rewrite [math]\sum_{n=1}^{\infty}n^2[/math] as [math] \frac{n(n+1)(2n+1)}{6}[/math] Now if you just take a quick glance at it, your power of n on top is 1, and your highest power of n on the bottom is 1.5 so when you take the limit to infinity, you will go to zero. If you want a more formal proof, there are a number of tests you could try, but it seems like it's been so long since calc 2, and I am not sure which one exactly would work. Link to comment Share on other sites More sharing options...

CPL.Luke Posted July 21, 2006 Author Share Posted July 21, 2006 neonblack, I don't think that I'm all that familiar with how you were able to re-write the sum, could you possibly elaborate on that? Link to comment Share on other sites More sharing options...

matt grime Posted July 21, 2006 Share Posted July 21, 2006 It is a well known identity: the sum of the first n squares. You can prove it by induction. It is in general true that the sum of the first r powers of n is a polynomial in n of degree r+1. Another, way to prove this result is to divide top and bottom by n, so we are investigating [math] \frac{1}{\sqrt{\sum\frac{1}{n}}}[/math] and that tends to zero since harmonic series [math]\sum \frac{1}{n}[/math] diverges. Link to comment Share on other sites More sharing options...

CPL.Luke Posted July 23, 2006 Author Share Posted July 23, 2006 could you also rewrite the series in the equation as 1/3n^3 - 1/3? ^getting the above by taking the integral from 1 -> n Link to comment Share on other sites More sharing options...

shmoe Posted July 24, 2006 Share Posted July 24, 2006 [MATH]\lim_{n\rightarrow \infty}\frac{n}{\sqrt{\sum_{n=1}^{\infty} n^2}}[/MATH] Do you mean: [MATH]\lim_{n\rightarrow \infty}\frac{n}{\sqrt{\sum_{k=1}^{n} k^2}}[/MATH] As you had it doesn't make much sense. It's not the case that [MATH]\sum_{k=1}^{n} k^2[/MATH] is equal to [MATH]\int_{1}^{n} t^2 dt[/MATH] either' date=' but you can use integrals to bound the sum: [MATH']\int_{0}^{n} t^2 dt\leq\sum_{k=1}^{n} k^2\leq\int_{1}^{n+1} t^2 dt[/MATH] Note the endpoints carefully (draw a graph of these). Your (n^3-1)/3 will be a lower bound also, good enough for this problem (maybe you just meant to bound it, but you had said "rewrite", which was troubling). See http://mathworld.wolfram.com/PowerSum.html for more on power sums, equation (23) and (33) are the relevant ones here. Link to comment Share on other sites More sharing options...

CPL.Luke Posted July 24, 2006 Author Share Posted July 24, 2006 shmoe good call on the mistake (although what is the real mathmatical difference , the equation is meant to give the angle between the vector [MATH]\sum_{k=1}^n k \vec{e_k}[/MATH] and [MATH]\vec{e_n}[/MATH] as the n dimensional space tends toward an infinite number of dimensions. Link to comment Share on other sites More sharing options...

matt grime Posted July 24, 2006 Share Posted July 24, 2006 There is a significant difference between the two things. The first does not actually make sense, now I think about it. Don't let laziness get in the way of doing things properly: these things can make serious differences. note, youe former vector, sum ke_k does not actually make much sense unless you want to talk about infinite dimensional vector spaces, and then you need to worry about what that infinite sum means (does it even exist?): it is certainly not an element of [math]\coprod_{\aleph_0} \mathbb{R}[/math] for instance. Link to comment Share on other sites More sharing options...

Neil9327 Posted July 24, 2006 Share Posted July 24, 2006 I'd have thought that a more common sense explanation is that it is equal to [math] \lim_{n\rightarrow \infty}\frac{n}{\sqrt{n^2+{\sum_{n=1}^{\infty} n^2}}} [/math] where this isn't a normal sum symbol - it goes from 1 to infinity but does not include the value n provided externally. So since this means that as n increases the bottom half will progressively get larger than the top, so at infinity the result will tend to zero. Link to comment Share on other sites More sharing options...

matt grime Posted July 24, 2006 Share Posted July 24, 2006 You are using n as a dummy variable in two different ways in one expression, and as shmoe as pointed out that is not allowed. Link to comment Share on other sites More sharing options...

CPL.Luke Posted July 24, 2006 Author Share Posted July 24, 2006 well the problem was given to me in a vector calc text book with the two vectors supplied the first question asked what the angle between the two vectors would be, using the definition of an angle in an n dimensional space the angle would be given by [MATH] \arccos{\left( \frac{\vec{e_n} \dot \sum_{k=1}^n k \vec {e_k}}{\left |{\vec{e_n}}\right| \left|{\sum{k=1}^n k\vec {e_k}\right|}}\right)}[/MATH] when evaluated this will give the the equation I gave initially (with the arccosine left out) and as you take n to infinity you can see what the angle does as the number of dimensions goes to infinity, since the equation converges to zero the angle becomes 90 degrees, which makes sense. matt the sum is a member of R^n as long as n is less than infinity right? and in a limit the number n is never actually infinity so the sum is a member of R^n Link to comment Share on other sites More sharing options...

matt grime Posted July 24, 2006 Share Posted July 24, 2006 I think I misread what you wrote, so ignore me. Apart from the bit where I back shmoe up about using dummy variables properly. Link to comment Share on other sites More sharing options...

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