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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Heck, does it even converge?

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh, haha didn't notice that \(n!\) would be even. Never mind...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks 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?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0how about \[\sum_{n\ge1}\frac{(1)^{1^{n}}}{n}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0lmao that is actually such a useless series now that i think about it

Empty
 one year ago
Best ResponseYou've already chosen the best response.1You 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
 one year ago
Best ResponseYou've already chosen the best response.0So two positive, one negative, is that the general pattern?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i want to say it converges

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Or 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
 one year ago
Best ResponseYou've already chosen the best response.1Ok 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
 one year ago
Best ResponseYou've already chosen the best response.0it feels like its right on the edge

Empty
 one year ago
Best ResponseYou've already chosen the best response.1The difference between 1 and .999... ? :P

Empty
 one year ago
Best ResponseYou've already chosen the best response.1That reminds me, the pseries 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
 one year ago
Best ResponseYou've already chosen the best response.0ya thats what i was thinking about

dan815
 one year ago
Best ResponseYou've already chosen the best response.0hmm how about rewriting it like this

dan815
 one year ago
Best ResponseYou've already chosen the best response.0maybe it might diverge actually

Empty
 one year ago
Best ResponseYou've already chosen the best response.1This 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
 one year ago
Best ResponseYou've already chosen the best response.0I'm okay with accepting that all math is connected in some way, no matter how obscurely :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This link gives an interesting approach with asymptotic notation: http://math.stackexchange.com/a/1400659/170231

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not the same series, but in the same vein of this "family".

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I 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 \[\sum \frac{(1)^n}{n} = \sum \frac{1}{2n}  \sum \frac{1}{2n1}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0right but thats fine, just because they both diverge doesnt mean it wont converge right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, \(\infty\infty\) isn't a determinate form.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0does this converge or diverge

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For \(n\ge1\), you get \(n^{1/n}\le n\), so it looks divergent.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ya apparently that thing is actually nearly one order more than a harmonic series

dan815
 one year ago
Best ResponseYou've already chosen the best response.0for all positive numbers more than 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.1Here'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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