# Sum $\sum \frac{1}{n}\not \in N$ [duplicate]

If $S_n$ denote sum of $n$ terms of H.P. $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ ..... , Then prove using summation of series that $S_n\not\in N$ $\forall \ n \in N$;

## marked as duplicate by Amzoti, Ittay Weiss, Emily, Gerry Myerson, MicahMar 24 '13 at 5:15

• check out this post – Coffee_Table Mar 24 '13 at 2:59
• What do you mean by your emphasis on 'using summation of series?' – davidlowryduda Mar 24 '13 at 3:37

For $\,n\ge 2\,$:

$$S_n:=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}=\frac{n!+\frac{n!}{2}+\frac{n!}{3}+\ldots+\frac{n!}{n}}{n!}$$

If we now choose $\,k\in\Bbb N \;\;s.t.\;\;2^k\mid n!\;,\;\;2^{k+1}\nmid n!\,$ , then all the summands in the last expression's numerator are even except one, namely

$$\frac{n!}{2^k}$$

and thus that expression's numerator is odd, whereas the denominator isn't...

• are u sure this would be accepted as a answer to the question: " using summation of series"? – ABC Mar 24 '13 at 3:23
• No, I am not since I don't know what "using summation of series" can possibly mean. – DonAntonio Mar 24 '13 at 3:36
• @DonAntonio I don't know what I'm missing-If n=7, 7!/4 = 1260, which is even... – Lucas Jan 18 '15 at 21:24
• but in this case, k =2 and k+1 =3, so that 4| 7! and 8|7! also. So that k does not satisfy the condition that $2^k | n!$ and $2^{k+1}\nmid n!$ – user73195 Oct 4 '15 at 13:26