Prove that the number $$\sum_{k=2}^{n}{1\over k}$$ is not an integer.


marked as duplicate by S.C.B., Milo Brandt, Blue, Shailesh, Semiclassical Mar 26 '16 at 3:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This is a duplicate, I think. $\endgroup$ – Chad Shin Mar 26 '16 at 3:21
  • $\begingroup$ What do you mean "duplicate"? $\endgroup$ – Theodoros Mpalis Mar 26 '16 at 3:22
  • 6
    $\begingroup$ math.stackexchange.com/questions/2746/… $\endgroup$ – dezdichado Mar 26 '16 at 3:22
  • $\begingroup$ @dezdichado thank you very much! $\endgroup$ – Theodoros Mpalis Mar 26 '16 at 3:23
  • 2
    $\begingroup$ I recommend you use the search function. It's not always helpful, though. It might help for you to browse some other questions that have been asked before in your free time. $\endgroup$ – S.C.B. Mar 26 '16 at 3:28

Suppose that $\sum_{i=2}^n \frac{1}{i}$ is an integer, $n>1$. By Bertrand's postulate, there exists a prime $\frac{n}{2}<p<n$. If we make the common denominator of the fraction, we get $$\sum_{i=2}^n \frac{1}{i} = \frac{1}{n!}\sum_{i=1}^n (2\times 3 \times ... \times (i-1) \times (i+1) \times ... \times n)$$.

Note that $p$ divides the denominator of the above fraction, but in the numerator all terms in the sum except the term $(1\times 2 \times ... \times (p-1) \times (p+1) \times ... \times n)$,and because we have assumed that $\frac{n}{2}<p<n$, this term does not divide $p$ because it can't contain a multiple of $p$. Hence it follows that $p$ does not divide the numerator of the fraction. Hence it follows that the fraction cannot be made an integer for any $n$.

  • 1
    $\begingroup$ Does the "$\times$" symbol is the same as"$\cdot$" ? $\endgroup$ – Theodoros Mpalis Mar 26 '16 at 3:29
  • $\begingroup$ Yes, both mean multiplication, but the difference is that $\cdot$ often is used for multiplication in groups and rings, while $\times$ is used for multiplying real numbers . By the way, hope you know about Bertrand's postulate. Erdos proved it at the age of nineteen. $\endgroup$ – астон вілла олоф мэллбэрг Mar 26 '16 at 3:37
  • $\begingroup$ Of course. But I thought "$\times$" was used for multiplication in groups and rings and "$\cdot$" for real numbers. Maybe it's oposite in my country... $\endgroup$ – Theodoros Mpalis Mar 26 '16 at 3:47
  • $\begingroup$ Your country Greece? Cyprus? Macedonia? All said and done, maybe it is the opposite, but the textbook I use is written by a man from the United Kingdom, while I myself have seen it in use in India. $\endgroup$ – астон вілла олоф мэллбэрг Mar 26 '16 at 5:05
  • $\begingroup$ I'm from Greece my friend $\endgroup$ – Theodoros Mpalis Mar 26 '16 at 5:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.