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Here is some sum that I can guess an answer but somehow lack of nice solution.

$$\displaystyle\sum _{k=0}^{n-2} (-1)^k\frac{1}{\sin\left(\frac{(2k+1)\pi}{4n-2}\right)}=\;?$$

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What is your guess? – user9413 Jul 25 '11 at 20:41
$2\lfloor\frac{n}{2}\rfloor$ – old Jul 25 '11 at 20:45
How did you arrive at it? I mean how did you guess it. – user9413 Jul 25 '11 at 20:46
By Mathematica program – old Jul 25 '11 at 20:51
may be this idea could work:… – user9413 Jul 25 '11 at 21:00
up vote 10 down vote accepted

As Eric Naslund has noted, we need to show that $$ \sum_{k=0}^{2n}\frac{(-1)^{k}}{\sin\left(\frac{2k+1}{2n+1}\frac{\pi}{2}\right)}=2n+1, $$ and this is, in turn, equivalent to $$ \sum_{k=0}^{2n}\frac{1}{\sin\left(\frac{4k+1}{2n+1}\frac{\pi}{2}\right)}=2n+1. $$ (Just add $\pi$ to the angles corresponding to odd values of $k$.)

The numbers $z_k = x_k+i y_k = \exp\left(i\frac{4k+1}{2n+1}\frac{\pi}{2}\right)$ for $0\le k \le 2n$ are exactly the distinct roots of the equation $z^{2n+1}=i$, so what we are summing are thus the reciprocals of the imaginary parts of these roots. That is, we want to show that $\sum_{k=0}^{2n} \frac{1}{y_k} = 2n+1$.

Now, if $z=x+iy$ satisfies $z^{2n+1}=i$, then $x^2+y^2=1$ and $1=\mathrm{Im} (x+iy)^{2n+1} = \sum_{m=0}^n \binom{2n+1}{2m+1} (-1)^m x^{2n-2m} y^{2m+1}$. Eliminating $x$ from these two equations gives $$ 0= -1 + \sum_{m=0}^n \binom{2n+1}{2m+1} (-1)^m (1-y^2)^{n-m} y^{2m+1} =: p(y). $$ The right-hand side is a polynomial of degree $2n+1$, so its roots are exactly the $2n+1$ distinct numbers $y_k$ above (and no others).

EDIT: The numbers $z_0,\dots,z_{2n}$ are evenly distributed on the unit circle; one of them is $(-1)^n i$, and the others appear in pairs $\pm x+iy$. Thus the $y_k$'s are not all distinct, and the struck out argument above doesn't work. However, it is clear that $\mathrm{Im} (x+iy)^{2n+1} \le 1$ if $x+iy$ lies on the unit circle, so $p(y) \le 0$ for $-1 \le y \le 1$. Therefore, the zeros of $p(y)$ in the open interval $-1 < y < 1$ must be of even multiplicity (since they are local maxima of $p$). We know $n$ distinct such zeros $y=y_k$, and since $p$ has degree $2n+1$ we conclude that these zeros have multiplicity exactly two, and together with the simple zero $y=(-1)^n$ they exhaust the zeros of $p$ (that is, there are no zeros outside of the interval $[-1,1]$. So the zeros of $p$, counted with multiplicity, are exactly $y=y_0,\dots,y_{2n}$.

Since the constant term is $-1$, the coefficient of the $y^1$ term is the sum of the reciprocals of the roots; this coefficient comes from the term for $m=0$ since all other terms contribute only higher powers of $y$, and it equals $\binom{2n+1}{1} = 2n+1$, just as we wanted to show. Done!

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+1, this is excellent! Very clever use of polynomial coefficients! – Eric Naslund Jul 28 '11 at 15:35
Note: I think there is a small typo in the last centered equation, it should be $(1-y^2)^{n-m}$ instead of $(1-y^2)^{2n-2m}$. – Eric Naslund Jul 28 '11 at 15:35
@Eric: Corrected. Thanks. – Hans Lundmark Jul 28 '11 at 16:07
Hmm, I see now that part of the argument is not quite correct, since the $y_k$'s are not distinct. I'll see if I can fix that later... – Hans Lundmark Jul 29 '11 at 8:44
Now the reasoning is watertight, I hope! – Hans Lundmark Jul 30 '11 at 8:29

This is not a solution, but cleans the question up nicely. I am ending for the day, and will think about it again tomorrow, but I thought I would post this comment for anyone that might want it.

Lets replace $n$ by $n+1$ (That is, increase it by $1$) so that we are looking at


Notice that the angle is always between $0$ and $\frac{\pi}{2}$. Then by the symmetry of the $\sin$ function

$$\sum_{k=0}^{2n}(-1)^{k}\frac{1}{\sin\left(\frac{(2k+1)}{(2n+1)}\frac{\pi}{2}\right)}=2L(n)+(-1)^{n}.\ \ \ \ \ \ \ \ \ \ (1)$$

The problem is equivalent to showing that

$$\sum_{k=0}^{2n}(-1)^{k}\frac{1}{\sin\left(\frac{(2k+1)}{(2n+1)}\frac{\pi}{2}\right)}=2n+1.\ \ \ \ \ \ \ \ \ \ (2)$$ In other words, the average is $1$.

Indeed, combining the above with $(1)$ gives $2n+1=2L(n)+(-1)^n,$ and hence $$L(n)=\frac{1}{2}\left(1-(-1)^{n+1}\right)+n=2\lfloor\frac{n+1}{2}\rfloor$$

as desired.

Remarks: Upon first glance $(2)$ seems much more tractable. It also says that the average of these values is exactly $1$, which is easier to understand intuitively. I think the next stage is to use the negative sign and another symmetry to get a sum over the entire circle, that is $k$ ranging from $0$ to $2n+1$.

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