# [SOLVED]re-writing a sum

#### dwsmith

##### Well-known member
Prove the formula
$$\sum_{k=1}^{n}e^{ik\theta} = \frac{e^{i\left(n+\frac{1}{2}\right)\theta}-e^{i\frac{\theta}{2}}}{2i\sin\frac{\theta}{2}}$$

I have a hint that says consider the expression $e^{i\left(n+\frac{1}{2}\right)\theta}-e^{i\left(n-\frac{1}{2}\right)\theta}$.
How can I get the second exponential in the numerator to have that expression?

#### Sudharaka

##### Well-known member
MHB Math Helper
Prove the formula
$$\sum_{k=1}^{n}e^{ik\theta} = \frac{e^{i\left(n+\frac{1}{2}\right)\theta}-e^{i\frac{\theta}{2}}}{2i\sin\frac{\theta}{2}}$$

I have a hint that says consider the expression $e^{i\left(n+\frac{1}{2}\right)\theta}-e^{i\left(n-\frac{1}{2}\right)\theta}$.
How can I get the second exponential in the numerator to have that expression?
Hi dwsmith,

We shall start from,

$\sum\limits_{k = 1}^ne^{ik\theta} = \frac{e^{i\theta}\left(1 - e^{in\theta}\right)}{1 - e^{i\theta}}$

With reference to your thread, Basics of Fourier Series we have obtained,

$e^{i\theta} - 1 = 2ie^{i\frac{\theta}{2}}\sin\frac{\theta}{2}$

Therefore,

\begin{eqnarray}

\sum\limits_{k = 1}^ne^{ik\theta} &=& \frac{e^{i\theta}\left(e^{in\theta}-1\right)}{2ie^{i\frac{\theta}{2}}\sin\frac{\theta}{2}}\\

&=& \frac{e^{\frac{i\theta}{2}}\left(e^{in\theta}-1\right)}{2i\sin\frac{\theta}{2}}\\

&=& \frac{e^{i\left(n + \frac{1}{2}\right)\theta}-e^{i\frac{\theta}{2}}}{2i\sin\frac{\theta}{2}}\\

\end{eqnarray}

$\therefore \sum\limits_{k = 1}^ne^{ik\theta}=\frac{e^{i\left(n + \frac{1}{2}\right)\theta}-e^{i\frac{\theta}{2}}}{2i\sin\frac{\theta}{2}}$

Kind Regards,
Sudharaka.

Last edited:

#### dwsmith

##### Well-known member
Hi dwsmith,

We shall start from,

$\sum\limits_{k = 1}^ne^{ik\theta} = \frac{e^{i\theta}\left(1 - e^{in\theta}\right)}{1 - e^{i\theta}}$

With reference to your thread, Basics of Fourier Series we have obtained,
Shouldn't the sum be
$$\sum\limits_{k = 1}^ne^{ik\theta} = \frac{\left(1 - e^{i(n+1)\theta}\right)}{1 - e^{i\theta}}$$
where did the $e^{i\theta}$ in the front come from?

#### Sudharaka

##### Well-known member
MHB Math Helper
Shouldn't the sum be
$$\sum\limits_{k = 1}^ne^{ik\theta} = \frac{\left(1 - e^{i(n+1)\theta}\right)}{1 - e^{i\theta}}$$
where did the $e^{i\theta}$ in the front come from?
Note that $$\sum\limits_{k = 1}^ne^{ik\theta}$$ is a geometric series with common ratio $$e^{i\theta}$$. So do you know the sum of the first $$n$$ terms of a geometric series?

#### dwsmith

##### Well-known member
Note that $$\sum\limits_{k = 1}^ne^{ik\theta}$$ is a geometric series with common ratio $$e^{i\theta}$$. So do you know the sum of the first $$n$$ terms of a geometric series?
$$s_N = 1 + e^{i\theta} + e^{2i\theta} + \cdots + e^{in\theta}$$
Multiple by $z$.
$$e^{i\theta}s_N = e^{i\theta} + e^{2i\theta} + e^{3i\theta} + \cdots + e^{i(n+1)\theta}$$
Next make the subtraction $s_N - e^{i\theta}s_N$.
So
$$s_n(1 - e^{i\theta}) = 1 - e^{i(n+1)\theta}\iff s_n = \frac{1 - e^{i(n+1)\theta}}{1 - e^{i\theta}}.$$

#### Sudharaka

##### Well-known member
MHB Math Helper
$$s_N = 1 + e^{i\theta} + e^{2i\theta} + \cdots + e^{in\theta}$$
Multiple by $z$.
$$e^{i\theta}s_N = e^{i\theta} + e^{2i\theta} + e^{3i\theta} + \cdots + e^{i(n+1)\theta}$$
Next make the subtraction $s_N - e^{i\theta}s_N$.
So
$$s_n(1 - e^{i\theta}) = 1 - e^{i(n+1)\theta}\iff s_n = \frac{1 - e^{i(n+1)\theta}}{1 - e^{i\theta}}.$$
Note that we are starting from $$k=1$$. So the first term should be, $$e^{i\theta}$$.