Prove that for any $n\ge 1$ $$ 1 + \frac{1}{3} + \frac{1}{5} + … + \frac{1}{2n+1} $$ is not an integer.
A hint would be greatly appreciated. Thanks, Payam
Asked
Active
Viewed 245 times
2
-
1Possible duplicate of Prove that $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$ is not an integer – Vinyl_cape_jawa May 22 '18 at 08:49
-
@Vinyl_coat_jawa The given link deals with a different sum. Please explain why you think this is a duplicate. – Peter May 22 '18 at 09:03
-
4It is slightly hidden, but the linked question is all the current OP needs, since $$1+\frac13+\ldots+\frac1{2n+1}=1+\frac12+\frac13+\ldots+\frac1{2n+1}-\frac12\left(1+\frac12+\frac13+\ldots+\frac1{n}\right)$$ – DonAntonio May 22 '18 at 09:06
-
3@– DonAntonio: It is possible to explain how the right side of above equation is not integer? If A and B is not integer there is no reason to A-1/2 B is not integer. – P.Ba May 22 '18 at 12:48
-
A simple proof, similar in spirit to the corresponding proof of $1+\frac12+\frac13+\frac14+\cdots$, is to observe that among the denominators there is a unique one divisible by a hgher power of three than the others. In other words, instead of using the non-archimedean 2-adic triangle inequality we use the 3-adic triangle inequality. – Jyrki Lahtonen Jan 16 '19 at 08:19