May 8th, 2015, 10:03 AM  #1 
Senior Member Joined: Aug 2014 From: United States Posts: 136 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: $\displaystyle\sum\limits_{n=1}^\infty \frac 1{n^2} = 1+\sum\limits_{n\in S}\left\{\frac{n^2} {n^21}\right\}$ Where $\{ x\}$ denotes the fractional part of $x$. I am just wondering if anyone can confirm that this is true. 
May 8th, 2015, 11:18 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,265 Thanks: 2434 Math Focus: Mainly analysis and algebra 
Well, ${n^2 \over n^2  1} = 1 + {1 \over n^2  1}$ so the fractional part is $1 \over n^2  1$.

May 8th, 2015, 12:17 PM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 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: $$ \sum_{n=2}^\infty\frac{1}{n^21}=\frac34 $$ 
May 9th, 2015, 01:57 PM  #4  
Senior Member Joined: Aug 2014 From: United States Posts: 136 Thanks: 21 Math Focus: Learning  Quote:
Quote:
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 $\displaystyle \sum\limits_{n=1}^\infty \frac{1} {n^2}=1+\sum\limits_{n\in S} \frac{1} {n^21}\approx 1+\displaystyle\sum\limits_{n=2}^\infty \frac 1{n^21}=1.75$ I don't know if the partial fraction decomposition of $\displaystyle\frac 1{n^21}=\frac{1} {2(n1)}\frac{1} {2(n+1)}$ makes the original sum look easier to sum as it does for $1+\displaystyle\sum\limits_{n=2}^\infty \frac 1{n^21}$ 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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