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 May 8th, 2015, 10:03 AM #1 Senior Member   Joined: Aug 2014 From: United States Posts: 137 Thanks: 21 Math Focus: Learning Does this work? Let $S$ be the set of integers greater than one that cannot be written in the form $a^b$ for integers $a$ and $b$ both greater than one. I seem to have got the following: $\displaystyle\sum\limits_{n=1}^\infty \frac 1{n^2} = 1+\sum\limits_{n\in S}\left\{\frac{n^2} {n^2-1}\right\}$ Where $\{ x\}$ denotes the fractional part of $x$. I am just wondering if anyone can confirm that this is true. Thanks from CRGreathouse
 May 8th, 2015, 11:18 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2455 Math Focus: Mainly analysis and algebra Well, ${n^2 \over n^2 - 1} = 1 + {1 \over n^2 - 1}$ so the fractional part is $1 \over n^2 - 1$. Thanks from neelmodi
 May 8th, 2015, 12:17 PM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms This seems to hold for an impressive number of digits, so I suspect something is going on. Probably it is true, though I can't prove it at the moment. Perhaps this will help someone: $$\sum_{n=2}^\infty\frac{1}{n^2-1}=\frac34$$ Thanks from neelmodi
May 9th, 2015, 01:57 PM   #4
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Quote:
 Originally Posted by v8archie Well, ${n^2 \over n^2 - 1} = 1 + {1 \over n^2 - 1}$ so the fractional part is $1 \over n^2 - 1$.
Yes, I noticed, but I suppose writing it in that way would yield a stronger expression.

Quote:
 Originally Posted by CRGreathouse This seems to hold for an impressive number of digits, so I suspect something is going on. Probably it is true, though I can't prove it at the moment.
Thanks for checking (by the way, how did you do that for enough digits? All I have is my graphing calculator or one on the computer and it seems to take a very long time to add these numbers one by one to get even the first two decimal places to match up).

My derivation was trivial in my opinion. Consider splitting the sum into an infinite sum of geometric series. For every series with common ratio 1/r^2, we sum up a part of this series. We do this for every r that is NOT of the form a^b to avoid over counting. We also take the fractional part of these expressions to eliminate the extra 1 that has been over counted each time (as v8archie pointed out). Finally, we add 1 back in at the end to compensate.

Weird coincidence: This form that I originally had it in looks somewhat similar to Euler's infinite product, except that his product was over the numbers in the set of prime, which cannot be written as $a\times b$ rather than a^b. (Although a^b also applies).

So we ultimately have $\displaystyle \sum\limits_{n=1}^\infty \frac{1} {n^2}=1+\sum\limits_{n\in S} \frac{1} {n^2-1}\approx 1+\displaystyle\sum\limits_{n=2}^\infty \frac 1{n^2-1}=1.75$

I don't know if the partial fraction decomposition of $\displaystyle\frac 1{n^2-1}=\frac{1} {2(n-1)}-\frac{1} {2(n+1)}$ makes the original sum look easier to sum as it does for $1+\displaystyle\sum\limits_{n=2}^\infty \frac 1{n^2-1}$

Also, I don't know if this is true or if it was a lie, but are there such "degrees" of primes, for example, the case where p cannot be written as $a\times b$ gives the "first degree" of primes. When p cannot be written as $a^b$ gives the second degree, etc?

Last edited by neelmodi; May 9th, 2015 at 02:01 PM.

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