Sums of reciprocals Suppose $n\ge2$ . Prove that neither 
the sum  $\sum_{i=2}^n \frac{1}{i}$, nor the sum $\sum_{i=2}^n \frac{1}{2i+1}$
is an integer
The approach a tried to take, was to show that the denominator is never a divisor of the numerator, but I can't figure out how to generalise this for any n
 A: You can get around Bertrand's postulate for the first sum by looking at powers of $2$: let $k$ be the greatest integer such that $2^k \leq n$. Then $2^k$ divides exactly one denominator of the series, since the least proper integer multiple of $2^k$ is $2^{k+1}$, which is greater than $n$. Therefore when you put the series over a common denominator, the denominator is divisible by $2$ while the numerator is of the form $2a + 1$.
EDIT: In fact the argument works for the second sum as well, by looking at $3^k$ instead of $2^k$. The first two multiples of $3^k$ are $2\cdot 3^k$,which is even and so not in the series, and $3^{k+1}$, which is greater than $n$.
A: Consider the largest prime $p \leq n$. When you express $\sum_{i=2}^n \frac{1}{i}$ as a numerator with $n - 1$ terms over the denominator $1 \cdot 2 \cdot \ldots \cdot n$, all but one of the terms in the numerator is divisible by $p$, hence the numerator is not divisible by $p$, but the denominator is divisible by $p$. Therefore, $\sum_{i=2}^n \frac{1}{i}$ can't be an integer.
The same argument applies to the second sum.
