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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 $S_n\not\in N$ $\forall \ n \in N$;

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

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

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  • $\begingroup$ are u sure this would be accepted as a answer to the question: " using summation of series"? $\endgroup$ – ABC Mar 24 '13 at 3:23
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    $\begingroup$ No, I am not since I don't know what "using summation of series" can possibly mean. $\endgroup$ – DonAntonio Mar 24 '13 at 3:36
  • $\begingroup$ @DonAntonio I don't know what I'm missing-If n=7, 7!/4 = 1260, which is even... $\endgroup$ – Lucas Jan 18 '15 at 21:24
  • $\begingroup$ 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!$ $\endgroup$ – user73195 Oct 4 '15 at 13:26

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