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,
Joseph