nicetry007 wrote: Let $ 2^{k}\;\leq \; n \;<\; 2^{k+1}$. It is to be noted that every number $ m \;\neq\; 2^{k}$ and $ m \;\leq\;n$ has a power of $ 2$ strictly smaller than $ k$. Suppose not. Let $ m \; = \; 2^{l}\cdot s$ where $ l \;\geq \;k$ and $ s$ is odd and $ s \;\geq\; 1$. We have $ m \; = \; 2^{l}\cdot s \;\leq \;n\;< \;2^{k+1}\;\Rightarrow s \;< \;2^{k+1-l}\;\leq \; 2\;$ ( as $ l \;\geq\; k$) $ \Rightarrow s = 1$ (as $ s$ is odd ) $ \Rightarrow m \; = \; 2^{l}\;<\; 2^{k+1}\;\Rightarrow \; l \;\leq \; k$. But $ l\;\geq \;k$. Hence, $ l \; = \; k \;\Rightarrow \; m \; = \; 2^{k}$, which is a contradiction as $ m \;\neq\; 2^{k}$. Thus, we can conclude that every $ m \; \neq \; 2^{k}$ and $ m \;\leq \; n$ has a power of $ 2$ strictly smaller than $ k$. The denominator of the sum is the $ lcm(2\;,\;3\;,\;\cdots\;,\;n)$ which has $ 2^{k}$ as a factor. Thus, when we take the sum of all the unit fractions, the numerator of every fraction other than $ \frac{1}{2^{k}}$ gets multiplied by a power of $ 2$ and the numerator of $ \frac{1}{2^{k}}$ gets multiplied by an odd number. Hence, the numerator of the sum of all the fractions is an odd number as $ odd+even \; = \; odd$. Therefore, the sum can never be an integer as the numerator is odd and the denominator is even.

Sorry to revive this but I remember there was a proof using Bertrand's postulate and factorials , but I can't remember the whole thing .Can anyone post it?

aham wrote: Sorry to revive this but I remember there was a proof using Bertrand's postulate and factorials , but I can't remember the whole thing .Can anyone post it? Consider $\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}=\frac{\frac{n!}{1}+\frac{n!}{2}+\cdots+\frac{n!}{n}}{n!}$ By Bertrand's postulate, there exists a prime $p$ such that $\lfloor{\frac{n}{2}}\rfloor <p<n$ If $2p\le n$, we have $\lfloor{\frac{n}{2}}\rfloor\:<p \le \frac{n}{2}$ which is a clear contradiction. So, $2p>n$ Or in other words, $p\nmid \frac{n!}{p}$ Now Note that $p|\frac{n!}{k} \forall k \in \mathbb{N}, 1\le k \le n, k\ne p$ So, ${p\nmid \frac{n!}{1}+\frac{n!}{2}+\cdots+\frac{n!}{n}}$ but $p|n!$ $\Longrightarrow \frac{\frac{n!}{1}+\frac{n!}{2}+\cdots+\frac{n!}{n}}{n!}$ is never an integer.

gouthamphilomath wrote: By Bertrand's postulate, there exists a prime $p$ such that $\left\lfloor{\frac{n}{2}}\right\rfloor <p<n$ Be careful! There is no prime $p$ such that $\left\lfloor{\frac{1}{2}}\right\rfloor <p<1$.

Let $p$ be the largest prime among 2,3,....,n. Then by Bertrand's postulate, $p<2n$. Then we factorize each $m\in [2,n]$ as a product of prime numbers: $m=p_1p_2\cdots p_r$ (here the $p_i$ may duplicate); then $p$ appears in the factorizations only once (when we factorize $p$ itself). Let $p_1,p_2,\cdots,p_k$ be all the prime numbers between 2 and n, excluding the aformentioned $p$. Then we multiply $\frac 12+\frac 13+\cdots+\frac 1n$ with $p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$, and we raise the powers $e_i$ large enough to make $\frac 1m p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ an integer for all $m\neq p$. But $p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}\frac 1p$ is not an integer! So comes the conclusion.

MeKnowsNothing wrote: gouthamphilomath wrote: By Bertrand's postulate, there exists a prime $p$ such that $\left\lfloor{\frac{n}{2}}\right\rfloor <p<n$ Be careful! There is no prime $p$ such that $\left\lfloor{\frac{1}{2}}\right\rfloor <p<1$. To apply Bertrand's postulate, $n \ge 7$ the others are trivial.

bobthesmartypants wrote: Suppose that the numerator when simplified is even. This means that the denominator is odd. This can only happen if no fraction in the original list of fractions that are added has an even denominator; this means that the list of fractions added must be a single fraction $\dfrac{1}{a}$ where $a$ is an odd number. However, the numerator would be $1$ which is odd, so we are done.

I proved that the numerator must be odd for any sum $\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots +\dfrac{1}{a_n}$ for consecutive integers $a_1,a_2,\cdots a_n$. However, I don't see how this can immediately be applied to this problem.

Well, we have to use nicetry007's proof for the denominator to be even?

By Bertrand's postulate,there exists a prime $p$ such that $\lfloor \frac{n}{2}\rfloor <p\le 2\lfloor \frac{n}{2}\rfloor\le n$.If $2p\le n$,then $\lfloor \frac{n}{2}\rfloor <p\le \frac{n}{2}$ which is absurd.Thus if $2\le k\le n$ and $k\neq p$,then $p\nmid k$.So ∃$a,b\in \mathbb N,p\nmid a$ s.t. $\sum_{k=2,k\neq p}^{n}\frac{1}{k}=\frac{b}{a}$. Then $\sum_{k=2}^{n}\frac{1}{k}=\frac{b}{a}+\frac{1}{p}=\frac{pb+a}{pa}$. Since $p\nmid pb+a$,$\frac{pb+a}{pa}$ isn't positive integer.Therefore the proof is completed.$\blacksquare$