How do you create an alternating series with the sign being the same twice in a row? I am working on a Taylor series question and I have created a series which alternates however, it does so in doubles. in other words it follows the following pattern: $x$, $x$, $-x$, $-x$, $x$, $x$,... 
My goal is to write this series using summation notation. 
I know that to produce a normal alternative series I simply use $(-1)^n$. However, I've found this significantly harder to create in summation notation because any manipulation to the $(-1)^n$ part of the sum only moves the position where the alternating series starts. Any suggestions are appreciated!
edit: please note that in my example, "$x$" is just an arbitrary function and each other $x$ does not equal the other $x$'s.
 A: It seems that you want to simulate the series $1, 1, -1, -1, 1, 1, -1, -1, ...$. These kinds of alternating series are usually easiest to do using $\sin$ or $\cos$ functions. (Although the formula $(-1)^{n(n-1)/2}$ works just as well, I think it's kind of interesting to see this series modeled with a trig function, so I'm going to do it anyway.)
Think about a $\cos$ function going through $(0, 1)$, $(1, 1)$, $(2, -1)$, $(3, -1)$, $(4, 1)$ and so on.
Where are the maximums and minimums? They're in between points with the same values, like $\frac{0+1} 2=0.5$ and $\frac{2+3} 2=2.5$.
Which is the maximum? It's the one in between the points of higher value, which is $0.5$. This represents our right shift since $\cos 0$ is the maximum.
What's the period? It's two times the difference between two consecutive extrema, or $2(2.5-0.5)=4$. The period is usually $2\pi$, so our coefficient of $x$ must be $\frac{2\pi} 4=\frac \pi 2$.
Thus, we have $\cos\left(\frac \pi 2\left(x-\frac 1 2\right)\right)$. However, this gives us $\frac{1}{\sqrt 2}$ for $x=0$, so we must multiply by $\sqrt 2$ to get the following function:
$$f(x)=\sqrt 2\cos\left(\frac \pi 2\left(x-\frac 1 2\right)\right)$$
If we put this into WolframAlpha, we clearly see that this gives us the sequence we wanted.
A: There is a very general way to attack the problem of writing a repeating sequence of coefficients.
The discrete Fourier transform of the sequence "$1,1,-1,-1$", of length $N=4$,  is $0,1+\mathrm{i},0,1-\mathrm{i}$".  This says that $$
    0 \mathrm{e}^{-2\pi \mathrm{i} \cdot 0 n /N} + (1+\mathrm{i}) \mathrm{e}^{-2\pi \mathrm{i} \cdot 1 n /N} + 0 \mathrm{e}^{-2\pi \mathrm{i} \cdot 2 n /N} + (1 - \mathrm{i}) \mathrm{e}^{-2\pi \mathrm{i} \cdot 3 n /N}
$$ is a constant multiple of your desired sequence.  ("Constant multiple" because there are constant premultipliers I have ignored to make the above more direct.)  That expression, in which we may ignore the two terms starting with zero, produces a sequence starting from $n=0$:  $$
    2, 2, -2, -2, 2, 2, -2, -2, \dots  \text{.}
$$  So, $$
\frac{(1+\mathrm{i}) \mathrm{e}^{-2\pi \mathrm{i} \cdot 1 n /4} + (1 - \mathrm{i}) \mathrm{e}^{-2\pi \mathrm{i} \cdot 3 n /4}}{2}
$$ does what you want.
Just as for $(-1)^n$, we may rewrite this using trigonometric functions.  $$
\left(\cos \left(\frac{\pi  n}{2}\right)-\sin \left(\frac{\pi n}{2}\right)\right) (\cos (\pi  n)-i \sin (\pi  n))
$$
Say, instead of starting at $n=0$, you want to start at $n=1$.  We cyclically rotate the sequence to get the term we want at the position where $n=1$, i.e., we consider the sequence $-1,1,1,-1$.  This gives transform $0,-1+\mathrm{i}, 0, -1-\mathrm{i}$.  The term to use in your sum is then $$
\frac{(-1+\mathrm{i}) \mathrm{e}^{-2\pi \mathrm{i} \cdot 1 n /4} + (-1 - \mathrm{i}) \mathrm{e}^{-2\pi \mathrm{i} \cdot 3 n /4}}{2}  \text{,}
$$ which is equivalent to $$
\left(\sin \left(\frac{\pi  n}{2}\right)+\cos \left(\frac{\pi n}{2}\right)\right) (-\cos (\pi  n)+i \sin (\pi  n))  \text{.}
$$
A: Just for kicks:
$$f(n)=\lim_{x\to n_-}\left\lfloor{\sin \frac{\pi x}{2}}\right\rfloor$$
This can be used to generalize sequences like:
$$\underbrace{1,1,1,\cdots,1}_{k \space1's},\underbrace{-1,-1,-1,\cdots,-1}_{k \space 1's}, 1\cdots $$
Can be represented as
$$f(n)=\lim_{x\to n_-} \left\lfloor{\sin \frac{\pi x}{k}}\right\rfloor$$
I thought that this might a bit cool for the OP.
A: $\frac{i^n+i^{-n}}2=\{1,0,-1,0,\dots\}$ has period $4$, so putting it together with itself shifted by one gives
$$
\{1,1,-1,-1,\dots\}=\overbrace{\frac{i^n+i^{-n}}2}^{1,0,-1,0,\dots}+\overbrace{\frac{i^{n-1}+i^{1-n}}2}^{0,1,0,-1,\dots}\tag{1}
$$

Using the series $-\log(1-x)=\sum\limits_{n=1}^\infty\frac{x^n}n$, we get
$$
\begin{align}
\frac11-\frac12-\frac13+\frac14+\cdots
&=\sum_{n=1}^\infty\left(\frac{i^n+i^{-n}}2+\frac{i^{n-1}+i^{1-n}}2\right)\frac1n\\
&=-\frac12\log(1-i)-\frac12\log(1+i)+\frac i2\log(1-i)-\frac i2\log(1+i)\\
&=-\frac12\log(2)+\frac i2\log(-i)\\[6pt]
&=\frac{\pi-\log(4)}4\tag{2}
\end{align}
$$
A: If you are looking for a formula for the sequence $1,1,-1,-1,1,1,-1,-1,\ldots$ then try some of the suggestions at OEIS A057077, such as $$(-1)^{n(n-1)/2}$$
A: Instead of trying to represent the sign sequence directly, you can also approach this problem by grouping you summands into pairs. So for a sum of the form
$f(0)+f(1)-f(2)-f(3)+f(4)+\dots$,
you would write 
$\sum_i(-1)^i(f(2i)+f(2i+1))$.
A: Here's a more "blunt force" way of doing this. Given a sequence like
$$1 + \frac12 - \frac13 -\frac14 + \frac15 + \frac16 - \frac17 - \frac18 + \cdots,$$
then you could notate it as follows:

$$\sum_{n=1}^\infty \frac{s(n)}{n},$$
  where
  $$s(n) = \begin{cases} 1 & \text{if } n \mod 4 = 1 \operatorname{or} 2, \\ -1 & \text{if } n \mod 4 = 3 \operatorname{or} 0\end{cases}$$

