anonymous
  • anonymous
Is there a closed form for the series \[\sum_{n\ge1}\frac{(-1)^{n!}}{n}\,?\]
Calculus1
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Heck, does it even converge?
Empty
  • 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
  • anonymous
Oh, haha didn't notice that \(n!\) would be even. Never mind...

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dan815
  • dan815
you're forgiven
anonymous
  • 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
  • Empty
Yessssssssssssss
dan815
  • dan815
how about \[\sum_{n\ge1}\frac{(-1)^{-1^{n}}}{n}\]
dan815
  • dan815
lmao that is actually such a useless series now that i think about it
Empty
  • 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
  • anonymous
So two positive, one negative, is that the general pattern?
Empty
  • Empty
Yeah, exactly.
dan815
  • dan815
i want to say it converges
Empty
  • 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
  • Empty
:P
Empty
  • 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
  • dan815
i think it will still
dan815
  • dan815
until infinity
dan815
  • dan815
it feels like its right on the edge
Empty
  • Empty
The difference between 1 and .999... ? :P
Empty
  • 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
  • dan815
ya thats what i was thinking about
dan815
  • dan815
hmm how about rewriting it like this
dan815
  • dan815
maybe it might diverge actually
dan815
  • dan815
|dw:1439858847534:dw|
Empty
  • 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
  • anonymous
I'm okay with accepting that all math is connected in some way, no matter how obscurely :)
dan815
  • dan815
|dw:1439858901052:dw|
anonymous
  • anonymous
This link gives an interesting approach with asymptotic notation: http://math.stackexchange.com/a/1400659/170231
anonymous
  • anonymous
Not the same series, but in the same vein of this "family".
dan815
  • dan815
|dw:1439859177697:dw|
Empty
  • 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
  • dan815
right but thats fine, just because they both diverge doesnt mean it wont converge right
anonymous
  • anonymous
Right, \(\infty-\infty\) isn't a determinate form.
Empty
  • Empty
|dw:1439859395039:dw|
dan815
  • dan815
|dw:1439859500069:dw|
dan815
  • dan815
does this converge or diverge
anonymous
  • anonymous
For \(n\ge1\), you get \(n^{1/n}\le n\), so it looks divergent.
dan815
  • dan815
ya apparently that thing is actually nearly one order more than a harmonic series
dan815
  • dan815
|dw:1439859913671:dw|
dan815
  • dan815
did u know that?
dan815
  • dan815
for all positive numbers more than 1
anonymous
  • 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
  • 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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