How to create alternating series with happening every two terms

I'm looking for a technique for creating alternating negatives and positives in a series. Specifically: when n=1, the answer is +, n=2 is +, n=3 is -, n=4 is -... etc.

I have every other part of the series written but I can't figure out that last piece... here's what I have now:

$$\sum_1^\infty 2^{n-1}(1^n+(-1)^n)/(3^{n-1}n!)*x^n$$

Technically, every other term is 0 so there doesn't really need to be two negatives in a row, it just has to sync up where I need them--I'm just guessing that I'd need it to work that way. Thanks for your assistance!

Here's a little trick which you might appreciate, or which you might find to obfuscate the matter. The triangular numbers $1,3,6,10,15,\dots$, given by the formula $T_n=n(n+1)/2$, have the property that $T_{4k+1}$ and $T_{4k+2}$ are odd and that $T_{4k+3}$ and $T_{4k+4}$ are even. So the expression $-(-1)^{T_n}=(-1)^{T_n+1} = (-1)^{(n^2+n+2)/2}$ alternates in sign in the way that you are looking for. In other words, if your original series is $\sum_{n=1}^\infty a_n$, where all $a_n$ are positive, then the modified alternating series you want is $$\sum_{n=1}^\infty (-1)^{(n^2+n+2)/2}a_n.$$

• Cool! $+1{}{}{}{}{}$ – user122283 Apr 25 '14 at 2:25
• This was fascinating. I know I'm not supposed to thank you here, but I really do appreciate this. It was illuminating! – Lauren Wolfe Apr 25 '14 at 18:23

$$S=\sum^\infty_{n=1}(-1)^{n+1}f(2n-1)+\sum^\infty_{n=1}(-1)^{n+1}f(2n)=f(1)+f(2)-f(3)-f(4)+...$$

• She wants it to be $f(1)+f(2)-f(3)-f(4)+f(5)+f(6)-\cdots$. That is, each sign repeats twice. (If I understand correctly) – apnorton Apr 25 '14 at 2:14
• Yes, @anorton is correct! It's a little trickier and I'm not quite sure how to go about it. – Lauren Wolfe Apr 25 '14 at 2:15
• @LaurenWolfe See my edit. – user122283 Apr 25 '14 at 2:20
• @anorton Thank you! :-) – user122283 Apr 25 '14 at 2:21

$$\sqrt 2 \; \sin \left( \frac{(2n-1)\pi}{4} \right)$$

Try: $$a_n = (-1)^{\frac{\left(2n+1+(-1)^{n+1}\right)}{4}+1}$$

Then, $$\langle a_n\rangle = 1, 1, -1, -1, 1, 1, \ldots$$

Derivation:
"Any sufficiently advanced technology is indistinguishable from magic." - Clark's Third Law
I don't really know how to describe how I got that... ;)

• When using powers, I think you should use \frac{blah}{blah}, not \dfrac{blah}{blah} - it becomes kind of hard to interpret as an exponent. – user122283 Apr 25 '14 at 2:22
• @SanathDevalapurkar I just made it a normal \frac, but it is a little harder to read. I'll leave it as-is. – apnorton Apr 25 '14 at 2:24

I must confess that I do not see the relation between your interesting questions and the expression you give. In practice,$$\sum_1^\infty 2^{n-1}(1^n+(-1)^n)/(3^{n-1}n!)*x^n=3 \left(\cosh \left(\frac{2 x}{3}\right)-1\right)$$ just involves even powers of $x$.