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Let $S=\sum_{k=1}^n \frac{1}{2k+1}$, how can we prove with elementary math reasoning that $S$ is not an integer? Can somebody help?

RicardoCruz
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    Are you allowed to use Bertrand's postulate? (Alternatively, clarify what elementary math reasoning refers to) – Calvin Lin Jul 07 '13 at 17:11
  • Hint: If $n$ is even, what do we know about the parity of the numerator and denominator? – A.E Jul 07 '13 at 17:13
  • @CalvinLin Bertrand's postulate is not allowed. Elementary: Prime factorization, gcd, lcm and divisibility, for example. – RicardoCruz Jul 07 '13 at 17:18

3 Answers3

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Hint: Recall the (elementary) proof that $\sum_{i=1}^n \frac{1}{i}$ is not an integer (for $n>1$):

Let $2^k$ be the largest power of 2 that is smaller than or equal to $n$. Consider making (smallest) common denominator of the form $2^k L$, where $L$ is odd. The numerator of every term will be even except for the term of $\frac{1}{2^k}$, which contributes an odd term $L$. Hence, the numerator is odd and the denominator is even. This cannot be an integer.

Hint: For finite sum of odd reciprocals, show that when we make a (smallest) common denominator, the numerator is not a multiple of $3$.

Solution: Let $3^k$ be the largest power of 3 that is smaller than or equal to $n$. Consider making common denominator of the form $3^k L$, where $L$ is not a multiple of 3. The numerator of every term will be a multiple of 3, except for terms of the form $\frac{ 1}{ a3^k}$ for some integer $a$.
Use the fact that $a=2$ does not appear in the sequence since it is even. Also, by the definition of $3^k$, no other higher multiple can appear. Hence, there is only 1 term which contributes a non-multiple of 3. Thus, the numerator is not a multiple of 3.

Calvin Lin
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Let $p=2k'+1$ be the largest odd prime number less than or equal to $2n+1$. Now consider the sum:

$$\frac13 + \frac15 + \frac17 + \cdots + \frac{1}{2k'-1} + \frac{1}{p} + \frac{1}{2k'+3} + \cdots + \frac{1}{2n+1}.$$

We can find a common denominator for this fraction by taking the product:

$$3\cdot5\cdots (2k'-1) \cdot p \cdot (2k' +3) \cdots (2n+1) = M.$$

In the numerator we have a sum of several terms:

$$\frac{M}{3} + \frac{M}{5} + \cdots + \frac{M}{p} + \cdots + \frac{M}{(2n+1)}.$$

Each term here is an integer.

I claim that all but $\frac{M}{p}$ is divisible by $p$. This would mean the numerator is not divisible by $p$ and is congruent to $\frac{M}{p}$ modulo $p$. Since $p$ appears in the denominator, this sum cannot be an integer.

Suppose that $\frac{M}{p}$ was indeed divisible by $p$. This means that we must have had another multiple of $p$ appear in one of our denominators. The next valid term would be $3p$. However, Bertrand's postulate tells us that for any natural number $n$ we have a prime between $n$ and $2n$. This tells us that there would have been a new largest prime between $p$ and $3p$. Which is a contradiction.

Joel
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  • I would like to comment that this is a standard trick in combinatorics and number theory. You always try to take the largest possible object and leverage that to prove your result. – Joel Jul 07 '13 at 17:17
  • How do you know that $3p$ doesn't appear in the denominator? This requires the use of Bertrand's postulate. – Calvin Lin Jul 07 '13 at 17:18
  • $3p$ does appear in the denominator (unreduced). That isn't a problem. We only are concerned with wether the numerator is divisible by $p$. – Joel Jul 07 '13 at 17:20
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    You said that all but $M/p$ is divisible by $p$, which is not true, if $M$ also includes $3p$, then $M/p$ has a factor of $3p$ , hence is a multiple of $p$. – Calvin Lin Jul 07 '13 at 17:22
  • You defined $p$ as the largest odd prime $\leq 2n+1$. Clearly $p$ is not a prime factor in $1,3,\ldots,p-2$. But how do you know it is not a prime factor in one of $p+2,p+4,\ldots,2n+1$? In other terms, how do you conclude that $p$ does not divide $M/p$ without invoking a more sophisticated argument? – Julien Jul 07 '13 at 17:25
  • Ah I see your point now. I suppose I would need Bertrand. – Joel Jul 07 '13 at 17:25
  • Alrighty. I added the last part of the argument. Though there is probably a simpler way that avoids Bertrand. Which is of course Calvin's proof haha. – Joel Jul 07 '13 at 17:35
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    @Joel Yes there is, read my solution. Note that OP requested no Bertrand. – Calvin Lin Jul 07 '13 at 17:36
  • This is true. I just didn't want to leave a hole in my argument. Nice work @Calvin. – Joel Jul 07 '13 at 17:39
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You can write

$$S=\frac{\sum_j\prod_{k\neq j}(2k+1)}{\prod (2k+1)}$$

Now the largest prime factor less or equal than $2n+1$ divides the denominator and all but one of the summands in the numerator, hence not the sum.

Simon Markett
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  • This requires the use of Bertrand, or at least that there is a prime between $n$ and $2n+1$. – Calvin Lin Jul 07 '13 at 17:17
  • You don't need Bertrand's postulate for this. You only need well ordering and the existence of prime numbers. – Joel Jul 07 '13 at 17:19
  • I should also add that you don't need a prime between $n$ and $2n+1$. Just need some largest prime less than $2n+1$. – Joel Jul 07 '13 at 17:21
  • @Joel No, conisder $ 1 + 1/3 + 1/5 + 1/7 + 1/9 = 1689/945$. The numerator is a multiple of 3. – Calvin Lin Jul 07 '13 at 17:24
  • But it isn't a multiple of 7, which is what the argument is claiming. – Joel Jul 07 '13 at 17:28
  • @Joel Precisely, this is why it needs Bertrand's postulate, which you were arguing against. This is the same as your argument; how do you know that $p$ and $3p$ are not in the sequence? – Calvin Lin Jul 07 '13 at 17:29
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    Right, that would be Bertrand's postulate. Because we would know then that there is a new largest prime between p and 2p. – Joel Jul 07 '13 at 17:31