Is there a closed form for the series
$$\sum_{n\ge1}\frac{(-1)^{n!}}{n}\,?$$

Question

Is there a closed form for the series
$$\sum_{n\ge1}\frac{(-1)^{n!}}{n}\,?$$

anonymous · Answer

Heck, does it even converge?

empty · Answer

I think it diverges because after n=2 we lose the interesting aspect of $ (-1)^{n!} = 1$ So it simplifies to basically the harmonic series.

anonymous · Answer

Oh, haha didn't notice that $n!$ would be even. Never mind...

dan815 · Answer

you're forgiven

anonymous · Answer

Thanks dan :P

Well I suppose a more general question might be in order. Something like, what conditions on $b_n$ are sufficient such that
$$\sum_{n\ge1}\frac{(-1)^{b_n}}{n}$$
converges?

empty · Answer

Yessssssssssssss

dan815 · Answer

how about 
$$\sum_{n\ge1}\frac{(-1)^{-1^{n}}}{n}$$

dan815 · Answer

lmao that is actually such a useless series now that i think about it

empty · Answer

You know what this is really quite fascinating @SithsAndGiggles I think we already know if it alternates it converges. But what about if we have:

$b_{3n} = 1$
$b_{3n+1} = 0$
$b_{3n+2}=0$

we would have the series:

$$1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots $$

Does this converge? :O

anonymous · Answer

So two positive, one negative, is that the general pattern?

empty · Answer

Yeah, exactly.

dan815 · Answer

i want to say it converges

empty · Answer

Or I could try to generalize the problem like you two have done and try to make the most complicated looking but really actually trivial version of the harmonic series.

empty · Answer

:P

empty · Answer

Ok dan if it converges then can we say it will converge for any value? Let's say we only make one value negative every 1,000 numbers?

dan815 · Answer

i think it will still

dan815 · Answer

until infinity

dan815 · Answer

it feels like its right on the edge

empty · Answer

The difference between 1 and .999... ? :P

empty · Answer

That reminds me, the p-series test as they called it was that if p > 1 this series converges:

$$ \sum_{n=1}^\infty \frac{1}{n^p}$$

So you may be on to something here.

dan815 · Answer

ya thats what i was thinking about

dan815 · Answer

hmm how about rewriting it like this

dan815 · Answer

maybe it might diverge actually

dan815 · Answer

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