...that is yet unsolved.

I joined almost at launch, and things were RADICALLY different back then in terms of the website's interface. One of the differences is a forums section (which was a pretty awesome feature, I don't know why anyone decided to take it down), and one of the threads contained a cool problem:

Diverges or Converges?

**\[\displaystyle \sum _{ n=0 }^{ \infty }{ \frac {(-1)^n \tau(2n+1) }{2n+1 } } ,\]**

where \(\tau(N)\) denotes the number of positive integer divisors of N.

I have the answer. I also have two solutions. However, what is missing is a proof of the answer.

This seems to be a hard-core Number Theory problem. So whoever likes these, you're more than welcome to try this problem!

I shall post the answers I have if necessary. But first, give it your own shot.

Good luck!

## Comments

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TopNewestIt sure seems to converge to 1.5708...., which looks a lot like \( \dfrac { \pi }{ 2 } \) ...., but I have no proof. It'd be fascinating if it really is that. – Michael Mendrin · 2 years, 11 months ago

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– Julian Poon · 2 years, 6 months ago

Im getting something close to \(\frac{1-\sqrt(5)}{2}\)Log in to reply

– John Muradeli · 2 years, 6 months ago

The Golden Ratio? Wow! You're almost there! 0.00002% error! A hint: the expression involves \(\pi\).Log in to reply

– Michael Mendrin · 2 years, 6 months ago

John, by all means, if the solution involves \(\pi\), please post it for sure tomorrow, which is 3/14/15! Don't miss this wonderful opportunity to do so on such a significant day. Then I'll reshare it tomorrow.Log in to reply

– John Muradeli · 2 years, 6 months ago

2morrow I get declined to MIT <.>Log in to reply

– Michael Mendrin · 2 years, 6 months ago

Tomorrow I'm going to lose the lottery? What are you saying?Log in to reply

Anyway, I've been looking for this notebook I had solutions in for about an hour, and it's gone! Dunno what happened to it, but there I had lots of other cool stuff and I hope to find it someday.

When I do, I'll post it up right away.

Cheers – John Muradeli · 2 years, 6 months ago

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– John Muradeli · 2 years, 11 months ago

Honestly, I did not expect this problem to blow up after two months. But just to line things up, no it does not converge to that. Good try though.Log in to reply

For example, 13 would have 2 divisors, while 12 would have 6. Yes or no? – Michael Mendrin · 2 years, 11 months ago

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Oh, and for the record: the solutions look MAD complicated! Like, alien complicated. – John Muradeli · 2 years, 11 months ago

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I was too distracted with another problem I was working on. So, finally, I've decided to tackle a few electricity problems, and I think I'm doing all right with those. Even though I still don't like it much. – Michael Mendrin · 2 years, 11 months ago

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alrighty.

LOL!!!

s

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– Michael Mendrin · 2 years, 11 months ago

You know, you really come up with the coolest images and GIFsLog in to reply

Dam I gotta write like a Nobel-Peace Prize quality comment to get an upvote from you.

And that makes it only so much precious ;p – John Muradeli · 2 years, 11 months ago

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– Michael Mendrin · 2 years, 11 months ago

Okay, okay, you've just been upvoted. This problem is kind of interesting, even though number theory isn't my thing either. I'll go sleep on it.Log in to reply

\[55555555555555555555555555555555555555555555555555555555555555555555555555555!\]

(I'll bet \(5$\) that you read that as a factorial ;)) – John Muradeli · 2 years, 11 months ago

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– Julian Poon · 2 years, 11 months ago

Dang.Log in to reply

Wait...that thread doesn't exist anymore? – Bogdan Simeonov · 2 years, 11 months ago

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– John Muradeli · 2 years, 11 months ago

It probably does but it's really old and I forgot what was the name of the thread so... I just posted it all over.Log in to reply

– Bogdan Simeonov · 2 years, 11 months ago

Do you have a link to the solutions?Log in to reply

– John Muradeli · 2 years, 11 months ago

Nah - hence my 2nd to last line in the note.Log in to reply

– Bogdan Simeonov · 2 years, 6 months ago

Please post the solutions, it's pi day after all!Log in to reply