1

Background : I have little experience writing proofs. I found this problem in the book Solving Mathematical Problems by Terrence Tao. I wish to prove that $n^{th}$ harmonic number is never an integer. In the book it says that Bertrand's Postulate ($\forall n$, where $n\in Z^+\;\;\exists$, a prime $p$, such that $n<p<2n$) is required to prove it, but I tried to prove it in a more elementary manner. I believe I have found required proof and wish to have it verified and critiqued.

Theorem : For every $n, q$, where $n,q \in Z^+, n\geq 2 $, it is true that $$\sum_{i=1}^n\frac1n\neq q$$

Proof. We will assume that the sum is equal to an integer, and reach a proof by contradiction with a counterexample.

Multiplying the individual terms by each other to make them common in denominator then adding them together results in the fraction. As per assumption, it should be equal to an integer $q$.

$$\frac{(2\times3 \times ... \times n)+(1\times3\times ... \times n)+ ...+(1 \times 2 \times ... \times (n-1))} {n!}=q$$ Thus, $$n!q= (2\times3 \times ... \times n)+(1\times3\times ... \times n)+ ...+(1 \times 2 \times ... \times (n-1))$$ Since, $n\geq2$, therefore $n!$ is divisible by $2$.
$$2\bigg(\frac{n!}{2}\bigg)q= (2\times3 \times ... \times n)+(1\times3\times ... \times n)+ ...+(1 \times 2 \times ... \times (n-1))$$

Therefore, if $q$ is an integer, the numerator must be even for all positive integer $n$. However, taking the counterexample of $n=3$ the numerator equals $11$, which is not an even integer. This is a contradiction. Therefore the negation of our assumption must be true, and there do not exist positive integers $q, n$ such that the $n^{th}$ harmonic number equals $q$, as desired.

$QED.$

Rohit Singh
  • 1,143
  • You can find mathjax tutorial here: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Aniruddha Deshmukh Nov 12 '18 at 04:35
  • You can also have that $1 \times 2 \times \cdots \times \left( n - 1 \right) = n!q - \left( \sum\limits_{i = 1}^{n - 1} \prod\limits_{j = 1 \ j \neq i}^n j \right)$ and hence $n | 1 \times 2 \times \cdots \times \left( n - 1 \right)$ which is a contradiction. Usually, for proving something by contradiction, we do not counter examples. Rather, we use a strong fallacy that arises due to our assumption which we then call "contradiction". – Aniruddha Deshmukh Nov 12 '18 at 04:41
  • 5
    You are just showing, Riley, that for $n=3$ you don't get an integer. What meant to be done is to show that no matter what $n$ is, you don't get an integer. And this can be done without Bertrand, by considering the highest power of two involved. See https://math.stackexchange.com/questions/2746/is-there-an-elementary-proof-that-sum-limits-k-1n-frac1k-is-never-an-int – Gerry Myerson Nov 12 '18 at 04:43
  • It is useful to consider the number of factors of $2$ in $\operatorname{lcm}(1,2,3,\ldots,n)$ and note that there is only one number in $1,2,3,\ldots,n$ that has that number of factors of $2$. – robjohn Nov 12 '18 at 07:34
  • Thank you for all your comments, and critiques. I understand the errors I have made now and hope to learn from them by considering your critiques. Also, the ord$2$ proof is really cool thanks for showing me it. –  Nov 12 '18 at 11:09
  • \sum gives $\sum$. Put superscripts and subscripts on it. E.g. \sum_{n=1}^{\infty}2^{-n} =1 gives $\sum_{n=1}^{\infty}2^{-n}=1.$ Similarly \prod for a product: $\prod$. – DanielWainfleet Nov 13 '18 at 02:49
  • 1
    There seems to be quite a big mistake with the body of the question. It should be $\sum_{i=1} \frac{1}{i}$, otherwise it is just weird. – ck1987pd May 25 '22 at 17:53

1 Answers1

0

You have your "for all " and "there exists" mixed up. $q$ depends on $n$ so call it $q_n$. And "the numerator" depends on $n$ too so call it $u_n.$ The 1st sentence of your last paragraph should say: "Therefore if there exists $n$ such that $q_n\in \Bbb Z$ then there exists $n$ such that $u_n$ is even."... NOT "for all $n$."

For $n\in \Bbb Z^+$ let $S(n)=\sum_{j=1}^n (1/j).$ Let $V(n)$ be the largest non-negative integer $j$ such that $2^j\leq n.$ That is, $2^{V(n)}\leq n<2^{1+V(n)}.$

Prove that if $S(n)=\frac {A_n}{B_n2^{V(n)}}$ where $A_n, B_n$ are odd positive integers then $S(n+1)=\frac {A_{n+1}}{B_{n+1}2^{V(n+1)}}$ where $A_{n+1}, B_{n+1}$ are odd positive integers. This implies that $S(n)\not \in \Bbb N$ for $n\geq 2$ because $n\geq 2\implies V(n)\geq 1.$

The proof requires some care. Split it into 2 cases:(i). When $n+1$ is a power of $2.$ (That is , $n+1=2^{V(n+1)}),$ and (ii). When $n+1$ is not a power of $2.$

DanielWainfleet
  • 57,985
  • "Bertrand's Postulate" is actually a theorem, first proven in the early 19th century. I dk if it has any bearing on this Q – DanielWainfleet Nov 13 '18 at 03:34
  • Thank you, your answer was very in depth. I unfortunately can't upvote because my reputation is too low. I will have a go at proving that. –  Nov 13 '18 at 04:27
  • I once looked at the highest power of $3$ that divides the denominator of $S(n)$ when $S(n)$ is in lowest terms and found that for some $n$, it is less than the highest power of $3$ not exceeding $n$. With powers of $2$ it is just an even/odd issue but for powers of $3$ it gets complicated. – DanielWainfleet Nov 13 '18 at 08:17
  • 1
    You may not be able to upvote, Riley, but you are able to "accept" the answer by clicking in the check mark next to it, if you are satisfied with the answer. – Gerry Myerson Nov 13 '18 at 10:10