This can be written as $\zeta(1)$ (don't be afraid if you never saw this, search for Zeta Function).
\[1+\dfrac12+\dfrac13+\ldots=\prod_{p\in\mathbb P}\dfrac{p}{p-1}\]
where $\mathbb P$ is the set of primes.
Why does this hold? Think using unique prime factorization.
Search found 592 matches
- Sat Oct 10, 2015 12:26 am
- Forum: Number Theory
- Topic: Sequence
- Replies: 3
- Views: 4711
- Sat Oct 10, 2015 12:23 am
- Forum: Number Theory
- Topic: #Number Theory
- Replies: 2
- Views: 4151
Re: #Number Theory
See the article here.
viewtopic.php?f=26&t=3422
viewtopic.php?f=26&t=3422
- Sat Oct 10, 2015 12:22 am
- Forum: Number Theory
- Topic: Thue's Lemma
- Replies: 0
- Views: 2852
Thue's Lemma
Here is an article on Thue's Lemma.
- Sat Oct 10, 2015 12:20 am
- Forum: Number Theory
- Topic: Good positive integers
- Replies: 1
- Views: 2942
Re: Good positive integers
Aha. That problem of mine! It's quite easy actually. But it can be made a bit hard with more restrictions. Have you solved it?
- Mon Sep 21, 2015 10:42 pm
- Forum: Introductions
- Topic: Hi Everyone!
- Replies: 1
- Views: 6755
Re: Hi Everyone!
You are most welcome!
- Mon Sep 14, 2015 11:52 pm
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies: 34
- Views: 31301
Re: ONTC Final Exam
now it should be correct
- Mon Sep 14, 2015 3:26 pm
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies: 34
- Views: 31301
Re: ONTC Final Exam
You have to prove it.tanmoy wrote:$2^{n}+1$ is odd.So,$n^{2}-1$ is odd.So,$n$ is even.Let $n=2k$.Then $2^{n}+1=(2^{k})^{2}+1$.So,every divisor of $2^{n}+1$ is of the form $4m+1$.
So,$n^{2}-1 \equiv 1 (mod 4)$.
Or,$n^{2} \equiv 2 (mod 4)$,which is impossible.
So there is no solution.
- Sun Sep 13, 2015 11:16 pm
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies: 34
- Views: 31301
Re: ONTC Final Exam
Try this: $n^2-1|2^n+1$
- Sat Sep 12, 2015 7:22 pm
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies: 34
- Views: 31301
Re: ONTC Final Exam
Ya, this is the way to do it.
- Mon Sep 07, 2015 1:18 am
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies: 34
- Views: 31301
Re: ONTC Final Exam
I posted the solutions of the first one in my new blog. But my laziness is getting the better of me again. When that passes, I will start writing again there. Stay tuned lol