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

1. anonymous

Heck, does it even converge?

2. 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.

3. anonymous

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

4. dan815

you're forgiven

5. 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?

6. Empty

Yessssssssssssss

7. dan815

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

8. dan815

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

9. 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

10. anonymous

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

11. Empty

Yeah, exactly.

12. dan815

i want to say it converges

13. 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.

14. Empty

:P

15. 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?

16. dan815

i think it will still

17. dan815

until infinity

18. dan815

it feels like its right on the edge

19. Empty

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

20. 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.

21. dan815

ya thats what i was thinking about

22. dan815

hmm how about rewriting it like this

23. dan815

maybe it might diverge actually

24. dan815

|dw:1439858847534:dw|

25. 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.

26. anonymous

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

27. dan815

|dw:1439858901052:dw|

28. anonymous

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

29. anonymous

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

30. dan815

|dw:1439859177697:dw|

31. 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 $\sum \frac{(-1)^n}{n} = \sum \frac{1}{2n} - \sum \frac{1}{2n-1}$

32. dan815

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

33. anonymous

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

34. Empty

|dw:1439859395039:dw|

35. dan815

|dw:1439859500069:dw|

36. dan815

does this converge or diverge

37. anonymous

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

38. dan815

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

39. dan815

|dw:1439859913671:dw|

40. dan815

did u know that?

41. dan815

for all positive numbers more than 1

42. 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 $$n\to\infty$$.

43. Empty

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