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

- anonymous

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

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- anonymous

Heck, does it even converge?

- Empty

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

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

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## More answers

- dan815

you're forgiven

- anonymous

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

Yessssssssssssss

- dan815

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

- dan815

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

- Empty

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

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

- Empty

Yeah, exactly.

- dan815

i want to say it converges

- Empty

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

:P

- Empty

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

i think it will still

- dan815

until infinity

- dan815

it feels like its right on the edge

- Empty

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

- Empty

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

ya thats what i was thinking about

- dan815

hmm how about rewriting it like this

- dan815

maybe it might diverge actually

- dan815

|dw:1439858847534:dw|

- Empty

This reminds me of something I was looking at earlier, the number of numbers relatively prime to a power of a prime number is given by the totient function,
\[\phi(p^n) = p^n * (1-\frac{1}{p}) \] So we can set up a density of relatively prime numbers always:
\[\rho(p)=\frac{\phi(p^n)}{p^n} = 1-\frac{1}{p} \]
So if there are infinite numbers, they will be relatively prime to any finite number:
\[\rho(\infty) = 1\]
Ok I don't think that actually relates but now I've typed it it's too late but to try to bring this back on track a bit more:
----
What patterns other than these very trivial ones of periodically flopping signs will result in convergence? What if instead we do:
\(b_n = 1 \) if \(p_n\) is prime
\(b_n=0\) otherwise.

- anonymous

I'm okay with accepting that all math is connected in some way, no matter how obscurely :)

- dan815

|dw:1439858901052:dw|

- anonymous

This link gives an interesting approach with asymptotic notation: http://math.stackexchange.com/a/1400659/170231

- anonymous

Not the same series, but in the same vein of this "family".

- dan815

|dw:1439859177697:dw|

- Empty

I don't know if we can rearrange infinite sums like this or not dan, for instance this convergent sum can be rearranged into two divergent sums, so idk
\[\sum \frac{(-1)^n}{n} = \sum \frac{1}{2n} - \sum \frac{1}{2n-1}\]

- dan815

right but thats fine, just because they both diverge doesnt mean it wont converge right

- anonymous

Right, \(\infty-\infty\) isn't a determinate form.

- Empty

|dw:1439859395039:dw|

- dan815

|dw:1439859500069:dw|

- dan815

does this converge or diverge

- anonymous

For \(n\ge1\), you get \(n^{1/n}\le n\), so it looks divergent.

- dan815

ya apparently that thing is actually nearly one order more than a harmonic series

- dan815

|dw:1439859913671:dw|

- dan815

did u know that?

- dan815

for all positive numbers more than 1

- anonymous

@Empty I don't think the series you gave (+, +, -, +, +, -, ...) converges. If \(H_n\) denotes the \(n\)th partial sum \(\sum\limits_{k\ge1}^n\frac{1}{k}\), then
\[\begin{matrix}
H_n&=&1&+&\dfrac{1}{2}&+&\dfrac{1}{3}&+&\cdots\\
\dfrac{2}{3}H_n&=&2\bigg(\dfrac{1}{3}&+&\dfrac{1}{6}&+&\dfrac{1}{9}&+&\cdots\bigg)
\end{matrix}\]
Subtracting gives your series,
\[\frac{1}{3}H_n=1+\frac{1}{2}-\frac{1}{3}+\cdots\]
and \(H_n\) diverges as \(n\to\infty\).

- Empty

Here's an interesting and related series:
\[\sum_{n=1}^\infty \frac{1}{4n} + \frac{1}{4n+1} - \frac{1}{4n+2} - \frac{1}{4n+3}\]

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