I'm having trouble finding this sum

[Jncik]

Homework Statement

find

$$\sum_{-N1}^{+N1}e^{-j\omega n}$$

Homework Equations

The Attempt at a Solution

Let $$\lambda = e^{-j\omega}$$

we have

$$\sum_{-N1}^{+N1}\lambda ^{n} = \sum_{-N1}^{-1}\lambda ^{n} + \sum_{0}^{+N1} \lambda ^{n}$$

for the first i have

$$S = \lambda ^{-N1} + \lambda ^{-N1+1} + \lambda ^{-N1+2} + ... + \lambda ^{-2} + \lambda ^{-1}$$

$$-\lambda S = -\lambda ^{-N1+1} - \lambda ^{-N1+2} - \lambda ^{-N1+3} - ... - \lambda ^{-1} - \lambda ^{0}$$

hence

$$S = \frac{\lambda ^{-N1} - 1}{1-\lambda }$$

for the second i have

$$S2 = \lambda^{0} + \lambda^{1} + ... + \lambda^{N1-1} + \lambda^{N1}$$
$$-\lambda S2 = -\lambda^{1} - \lambda^{2} - ... - \lambda^{N1} - \lambda^{N1+1}$$

hence

$$S2 = \frac{1 - \lambda^{N1+1}}{1-\lambda}$$

so the sum is

$$\frac{1-\lambda^{N1+1} + \lambda^{-N1} - 1}{1-\lambda} = \frac{\lambda^{-N1} - \lambda^{N1+1}}{1-\lambda} = \frac{e^{j \omega N1} - e^{-j \omega N1}e^{-j\omega}}{1-e^{-j\omega}}$$

but the book says $$\frac{sin\omega(N1 + \frac{1}{2})}{sin(\frac{\omega}{2})}$$

 

[Dick]

Multiply numerator and denominator by e^(j*w/2). Now remember e^(jx)-e^(-jx)=2jsin(x).

 

[Jncik]

thanks a lot :)