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Note: i'm re-writing some of this to reflect some advice given below

I have reason to believe that this series:

$$ d = \lim_{n\to \infty }\sqrt{{\left({\left(2 \, n^{2} - n\right)} e^{\left(2 i \, \sqrt{n^{2} - n} \pi\right)} + {\left(n e^{\left(4 i \, \sqrt{n^{2} - n} \pi\right)} + n\right)} \sqrt{n^{2} - n}\right)} e^{\left(-2 i \, \sqrt{n^{2} - n} \pi\right)}} $$

definitely converges for n considered as an integer only, and I've calculated values for it up to 64K (Excel) without it going below .9

It represents a distance between 2 coordinates which approach some distance the larger n gets. I'm trying to find that limit so that I know how close the points ultimately get.

I think I've figured out how to take the limit, and doing so it looks like it goes to 0. I'm hoping someone can check my work and see if I have this right, and if not, where I've gone wrong.

First I avoid dealing with the square root and substitute: $$ a = \sqrt{n^{2} -n} $$

into the above so it's easier to look at (for me).

$$ d = \sqrt{a n e^{\left(-2 i \, \pi a\right)} + a n e^{\left(2 i \, \pi a\right)} + 2 \, n^{2} - n} $$

and then:

$$ d = \sqrt{2 \, a n \cos\left(-2 \, \pi a\right) + 2 \, n^{2} - n} $$

knowing that: $$ \lim_{n\to \infty } \sqrt{n^{2} -n} = n-1/2 $$

I substitute for a:

$$ d = \sqrt{2 \,n \left( n-1/2 \right) \cos\left(-2 \, \pi \left( n-1/2 \right)\right) + 2 \, n^{2} - n} $$

which reduces to:

$$ d = \sqrt{ \left( 2 \, n^{2} - n \right) \cos\left(- \pi n \right) + 2 \, n^{2} - n} $$

and then:

$$ d = \sqrt{ \left( 2 \, n^{2} - n \right) \left( \cos\left(- \pi n \right) + 1\right)} $$

and finally:

$$ d = \sqrt{ \left( 2 \, n^{2} - n \right) \left( -1 + 1\right)} = 0 $$

so: $$ d = \lim_{n\to \infty }\sqrt{{\left({\left(2 \, n^{2} - n\right)} e^{\left(2 i \, \sqrt{n^{2} - n} \pi\right)} + {\left(n e^{\left(4 i \, \sqrt{n^{2} - n} \pi\right)} + n\right)} \sqrt{n^{2} - n}\right)} e^{\left(-2 i \, \sqrt{n^{2} - n} \pi\right)}} = 0 $$

for (integer) n, which is definitely a nice answer, but have i missed something here?

Obviously I have, the answer given below of

$$ \sqrt{4+\pi^2}/4 $$

definitely looks much more like the values I've calculated.

Thanks in advance,


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I think you need to make more explicit what you're deriving from what -- you lost me at the point where it says "so ... and then". It's clearer if you write equations rather than just expressions that leave the reader wondering what these are supposed to be equal to. (You called the displayed expression at the very beginning an "equation", but it isn't.) From what I can tell in the current state of the question, there's a mistake: $\lim_{n\to \infty } \sqrt{n^{2} -n} = n-1/2$ doesn't make sense because the variable that's taken to a limit on the left-hand side can't occur in the result. –  joriki Aug 4 '11 at 15:45
I second what joriki said and add that $\cos(-\pi n) = -1$ only when $n$ is odd, so unless I'm missing something, that step is wrong as well. –  Jason DeVito Aug 4 '11 at 16:04
Like @joriki said. And the limit is not zero but $\frac14\sqrt{4+\pi^2}=0.931047945$. –  Did Aug 4 '11 at 16:50
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1 Answer

I would suggest that you introduce the auxiliary quantities $$z_n:=\exp\Bigl(2i\pi n\sqrt{1-{1\over n}}\Bigr)\ .$$ Using the $z_n$ your expression (without the outermost square root) looks like $$Q_n:=\Bigl((2n^2-n) z_n +n^2(z_n^2+1)\sqrt{1-{1\over n}}\Bigr)\ z_n^{-1}=2n^2-n+2n^2{z_n+z_n^{-1}\over 2}\sqrt{1-{1\over n}}\ .$$ In the next step we have to develop $\sqrt{1-{1\over n}}$ into powers of ${1\over n}$. Using the Taylor expansion of $\sqrt{1-x}$ at $0$ we get $$\sqrt{1-{1\over n}}=1-{1\over 2n}-{1\over 8n^2}+O(n^{-3})\ .$$ Now we look at $$\eqalign{{z_n+z_n^{-1}\over 2}&=\cos\Bigl(2\pi n \sqrt{1-{1\over n}}\Bigr)=\cos\Bigl(2\pi n\Bigl(1-{1\over 2n}-{1\over 8n^2}+O(n^{-3})\Bigr)\Bigr)\cr &=-\cos\Bigl({\pi\over 4n}+O(n^{-2})\Bigr)=-1+{\pi^2\over 32 n^2}+O(n^{-3})\ .\cr}$$ Here we have used the Taylor expansion of $\cos$ at $0$. Incidentally it has become evident that the $Q_n$ are in fact real. They can now be written as follows: $$\eqalign{Q_n &= 2n^2-n +2n^2\Bigl(-1+{\pi^2\over 32 n^2}+O(n^{-4})\Bigr)\bigl(1-{1\over 2n}-{1\over 8n^2}+O(n^{-3})\bigr)\cr &= {1\over 4}+{\pi^2\over16}+O(n^{-1})\ .\cr}$$ Therefore $$\lim_{n\to\infty}\sqrt{Q_n}={1\over4}\sqrt{4+\pi^2}\ ,$$ as stated by Didier Piau.

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Thanks for your help appreciate all I can get....I'm sure it's obvious that I'm no pro, but I have come across this series and would like to find the limit. Please forgive my misuse of terminology, I'm sure it grates. BTW, can i ask how you found the limit to be equal to $$\sqrt{4+\pi^2}/4$$ ? –  Joseph Anderson Aug 4 '11 at 21:42
Christian: when expanding the cosine, I believe the error term should be $O(n^{-3})$ and not $O(n^{-4})$. This has no effect on the validity of the subsequent steps. –  Did Aug 5 '11 at 9:49
@Didier Piau: Thank you. I've corrected that. –  Christian Blatter Aug 5 '11 at 12:02
Thank you again so much, i really appreciate your help –  Joseph Anderson Aug 5 '11 at 13:40
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