Looking for the limit of a complex exponential 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
 A: 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.
