How can we show 1 + 1/2 +1/3 +1/n......is never an integer if n >=2?

Question

More precisely, show that, for every n >=2, 1 + 1/2 +....1/n is not an integer.

Thank you for any help.

Answer 1

Assume the sum is an integer, call it k. We assume that k exists, for a contradiction.

1 + 1/2 + 1/3 + ... + 1/n = k

multiply by n! on both sides:

n! / 1 + n! / 2 + n! / 3 + .... + n! / n = n! k

Now, notice what happens if we take p to be the largest prime p≤n. Now subtract over:

n! / p = n! k - n!/1 - n!/2 - .... - n!/(p-1) - n!/(p+1) - ... - n!/n

These are all integers, and so divisibility must work out sensibly. On the right, everything is divisible by p.

The quantity n! / p is not, however. Thus the quantity is both divisible and not divisible by p. This is impossible, and thus k cannot be an integer after all.

Answer 2

let s(n) = 1 + 1/2.....+ 1/n. If n=1, S(1) = 1 is integer. For n=2, s(n) = 1/2 and for n= 3, 1 < s(n) < 2, cause 1/3 < 1/2. So, for n=1 and n=2 s(n) is not an integer.

For n >=4, write s(n) = (n!/1 + n!/2.....n!/n)/(n!). Let k be the largest positive integer so that 2^k <= n, that is, 2^k is the largest power of 2 less or equal n. If p is the exponent of 2 in the factorization of n! in prime numbers, then 2 <=
In the factorization in prime numbers of the elements of {1, 2,....n}, the number k, as exponent of 2, appears precisely in 2^k. If it appeared in some other m of this set, then, since m <> 2^k, we'd have m >= 3 * 2^k > 2 * 2^k = 2^(k+1) >n, contrarily to the fact that m is in {1, 2,....n}. Therefore, with exception of 2^k, all the numbers of this set have the prime factor 2 with exponent < k. This implies that, for each m =1, 2,...n, m <> 2^k, (n!)/m contains the prime factor 2 with exponent > p -k >0 and (n!)/(2^k) contains the factor 2 with exponent p - k.

Since s(n) n = (n!/1 + n!/2.....n!/n)/(n!), if we divide all the terms in the numerator by 2^(p-k), then, with exception of ((n!)/^(2^k), which gives an odd number, all the other n- 1 terms are even. So, the sum in the numerator, after the division, is odd. And since (n!)/(2^(p-k)) is even, because p > p - k, t follows s(n) is given by the ratio of an odd and an even number. So, s(n) is not an integer, completing the proof.

Answer 3

that's of course actual, simply by fact the countless sequence a million + a million/a million! + a million/2! + ... = e = 2.718 approx. Your sequence is monotonically increasing as n will boost, and it has a decrease sure of a million.5 and an bigger sure of e-a million < a million.8 although that isn't be a info, until you tutor some result like e < 3 this is a info: a million + a million/2! + a million/3! + ... + a million/n! <= a million + a million/2 + a million/2^2 + a million/2^3 + ... + a million/2^n this could be a geometrical sequence with sum = (a million - a million/2^(n+a million)) / (a million - a million/2) = 2 (a million - a million/2^(n+a million)) < 2) So the sum of the sequence is > a million and < 2 if n > a million