Thanks, SOS. Is there any problem you can't eventually solve?

:)

Anyway, my attempt was to use the exponential for sin. I chipped away at this off over the last several days. I think I made some headway.

Using what is inside the parentheses:

\(\displaystyle \sum_{k=1}^{n}\sin(k/n)=\frac{1}{2i}\sum_{k=1}^{n}\left(e^{\frac{ki}{n}}-e^{\frac{-ki}{n}}\right)\)

\(\displaystyle \sum_{k=1}^{n}\sin(\frac{2k-1}{2n})=\frac{1}{2i}\sum_{k=1}^{n}\left(e^{\frac{(2k-1)i}{2n}}-e^{\frac{-(2k-1)i}{2n}}\right)\)

So, we get:

\(\displaystyle \frac{\frac{1}{2i}\sum_{k=1}^{n}\left(e^{\frac{ki}{n}}-e^{\frac{-ki}{n}}\right)}{\frac{1}{2i}\sum_{k=1}^{n}\left(e^{\frac{(2k-1)i}{2n}}-e^{\frac{-(2k-1)i}{2n}}\right)}\)

Now, factor a little:

\(\displaystyle \frac{\sum_{k=1}^{n}\left(e^{i/n}\right)^{k}-\sum_{k=1}^{n}\left(e^{-i/n}\right)^{k}}{e^{-i/2n}\sum_{k=1}^{n}\left(e^{i/n}\right)^{k}-e^{i/2n}\sum_{k=1}^{n}\left(e^{-i/2n}\right)^{k}}\)

These are partial geometric series. They should simplify down.

The top left one evaluates to \(\displaystyle \frac{(e^{i}-1)e^{i/n}}{e^{i/n}-1}\)

The top right one: \(\displaystyle \frac{(e^{i}-1)e^{-i}}{e^{i/n}-1}\)

The bottom left: \(\displaystyle \frac{e^{-i/2n}(e^{i}-1)e^{i/n}}{e^{i/n}-1}\)

The bottom right: \(\displaystyle \frac{e^{i/2n}(e^{i}-1)e^{-i}}{e^{i/n}-1}\)

Putting these altogether, it should whittle down to some trig functions involving sin and/or cos. But, I have not finished yet.

It looks encouraging though.

EDIT:

Well, I spent some time trying to hammer down the results of the sums above.

They become:

\(\displaystyle \frac{\sin(1/n+1)-\sin(1/n)-\sin(1)}{\sin(1/2n+1)+\sin(1/2n-1)-2\sin(1/2n)}\)

this is equivalent to:

\(\displaystyle \frac{\cos(1/2n)\sin(1/2)+\sin(1/2n)\cos(1/2)}{\sin(1/2)}\)

Using the product-to-sum formulas on the numerator, it whittles down to:

\(\displaystyle =\frac{\sin(\frac{n+1}{2n})}{\sin(1/2)}\)

But, I admit I left tech do most of the work and played around with some trial and error.

So, we finally get:

\(\displaystyle \lim_{n\to \infty}\left(\frac{\sin(\frac{n+1}{2n})}{\sin(1/2)}\right)^{n}\)

Now, since this is a limit and there is an 'e' in the required solution, I figured I would make the sub \(\displaystyle n=1/k\) in order to get something that resembles the 'e' limit.

\(\displaystyle \lim_{k\to 0}\left(\frac{\sin(\frac{k+1}{2})}{\sin(1/2)}\right)^{1/k}\)

So, take logs:

\(\displaystyle \lim_{k\to 0}\frac{1}{k}[log(\sin(\frac{k+1}{2})-log(\sin(1/2))]\)

Using L'Hopital and taking this limit:

\(\displaystyle \lim_{k\to 0}\frac{1}{2}\cot(\frac{k+1}{2})\)

results in \(\displaystyle 1/2\cot(1/2)\)

Now, e:

\(\displaystyle e^{1/2\cot(1/2)}=e^{\frac{1}{2\tan(1/2)}}\)

WHEW!!!!. I messed around with this hoping it would lead somewhere eventually. The worst part, for me anyway, was finding a workable trig identity out of all of that.