Weird number theory sum.

Question

kainui · Answer

$$\tau(n) = \text{number of divisors of n}$$
So because 6 has exactly 1, 2, 3, and 6 as its divisors, this means:
$$\tau(6)=4$$
---
So my question is, does this converge to some finite value and if it does, what is it?

$$\frac{1}{\tau(1)^1} + \frac{1}{\tau(2)^2} + \frac{1}{\tau(3)^3}+\frac{1}{\tau(4)^4} +\cdots$$

superteenblue · Answer

I think you can do this yourself

princessstyles-horan · Answer

wow my mind is BLOWN

ganeshie8 · Answer

1/2^n maybe negligible, so it may converge

ganeshie8 · Answer

need to think a bit more thoroughly

kainui · Answer

I think it converges because:

$$\tau(n)^n \ge 2^n$$ for n>1 so we can compare the terms to the geometric series:

$$\frac{1}{2^n} \ge \frac{1}{\tau(n)^n}$$

Finding a closed form or close approximation to this might be difficult though.

kainui · Answer

Starting from here at what we know: $$\sum_{n=2}^\infty \frac{1}{2^n} \ge \sum_{n=2}^\infty \frac{1}{\tau(n)^n}$$

Then I "complete" the sums by moving their indexes to what is most convenient and subtracting the terms out:

 $$-1-\frac{1}{2}+\sum_{n=0}^\infty \frac{1}{2^n} \ge -1 + \sum_{n=1}^\infty \frac{1}{\tau(n)^n}$$

add 1 to both sides and simplify the geometric series:

$$\frac{-1}{2} + \frac{1}{1-\frac{1}{2}} \ge \sum_{n=1}^\infty \frac{1}{\tau(n)^n}$$
in other words:

$$\frac{3}{2}  \ge \sum_{n=1}^\infty \frac{1}{\tau(n)^n}$$

I think I can get a lower bound on it as well, which would be nice, or maybe by playing around with some geometric series we can decrease the upper bound too?

kainui · Answer

Conjecture:

$$\frac{199 \pi}{438} = \sum_{n=1}^\infty \frac{1}{\tau(n)^n}$$

thewafflebro · Answer

my mind cant handle this

kainui · Answer

Ok that's a totally bogus conjecture, I just did the sum up to a little over a thousand terms and the answer I got was
1.42734 
which is roughly that fraction haha.

It's not too far away from $\sqrt{2}$ but it's probably just a coincidence.

kainui · Answer

Playing around, apparently this is quite close: $$\sum_{n=1}^\infty \frac{1}{\tau(n)^n} \approx \sqrt{2} + \frac{5 \Omega_U}{3}$$ Where $\Omega_U$ is called http://mathworld.wolfram.com/ChaitinsConstant.html which is defined as: $$\Omega_U \equiv \sum_{p \ halts} 2^{-|p|}$$ Whatever that means, but hey, it looks similar so that's mildly promising and unexpected and probably wrong.