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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

user61067
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2 Answers2

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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 $n$.

user7530
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    In fact, the argument works for $\sum_{i=1}^n \frac{1}{mi+1}$ for any positive integer $m$ for which $m+1$ is a prime or prime power, by looking at $(m+1)^k$. – Robert Israel Feb 07 '13 at 19:42
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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.

ferson2020
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    Just for clarity: The fact that such $p$ occurs only once in the denominator requires Betrand's postulate - that there must be a prime between $n/2$ and $n$. – Thomas Andrews Feb 07 '13 at 18:49
  • I'm sure the answer answer is great too, but this one made more sense to me. Thank you :) – user61067 Feb 12 '13 at 23:13
  • @ferson2020 how would I prove that p divides all but one of the terms in the numerator? – user61067 Feb 12 '13 at 23:25
  • Each of the $n$ terms in the numerator looks like $2 \cdot 2 \cdot \ldots \cdot n$, but missing one of the terms. If $p$ is the largest prime less than $n$, and it is between $n/2$ and $n$, it doesn't divide any other number between $2$ and $n$. Therefore, $p$ doesn't divide the term in the numerator missing $p$ in the product, but every other term has a $p$ in the product, so $p$ divides every other term. – ferson2020 Feb 13 '13 at 15:28
  • If you want to avoid the messiness of making sure $p$ doesn't divide any other number between $2$ and $n$ (the Betrand's postulate part), consider the largest prime power $p^k$ instead. I think it's easier to see $p^k$ doesn't divide any other number between $2$ and $n$. – ferson2020 Feb 13 '13 at 15:30