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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!

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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 $$ \sum_{n=1}^\infty (-1)^{(n^2+n+2)/2}a_n. $$

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    $\begingroup$ Cool! $+1{}{}{}{}{}$ $\endgroup$ – user122283 Apr 25 '14 at 2:25
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    $\begingroup$ This was fascinating. I know I'm not supposed to thank you here, but I really do appreciate this. It was illuminating! $\endgroup$ – Lauren Wolfe Apr 25 '14 at 18:23
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$$S=\sum^\infty_{n=1}(-1)^{n+1}f(2n-1)+\sum^\infty_{n=1}(-1)^{n+1}f(2n)=f(1)+f(2)-f(3)-f(4)+...$$

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  • $\begingroup$ 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) $\endgroup$ – apnorton Apr 25 '14 at 2:14
  • $\begingroup$ Yes, @anorton is correct! It's a little trickier and I'm not quite sure how to go about it. $\endgroup$ – Lauren Wolfe Apr 25 '14 at 2:15
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    $\begingroup$ @LaurenWolfe See my edit. $\endgroup$ – user122283 Apr 25 '14 at 2:20
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    $\begingroup$ @anorton Thank you! :-) $\endgroup$ – user122283 Apr 25 '14 at 2:21
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$$ \sqrt 2 \; \sin \left( \frac{(2n-1)\pi}{4} \right) $$

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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... ;)

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  • $\begingroup$ 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. $\endgroup$ – user122283 Apr 25 '14 at 2:22
  • $\begingroup$ @SanathDevalapurkar I just made it a normal \frac, but it is a little harder to read. I'll leave it as-is. $\endgroup$ – apnorton Apr 25 '14 at 2:24
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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 $x$.

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