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Aryth · #17 February 6th, 2008, 02:43 PM

The Sum:

$\sum_{i = 1}^{k}\frac{1}{k}$

Does not yield an integer for all $k \geq 2$ .

For k = 2:

$1 + \frac{1}{2} = \frac{3}{2}$

That is not an integer.

Since the series is divergent, then we must reduce it:

$\left[\frac{(3*4*...*k) + (2*4*...*k) + ... + (2*3*...*k-1)}{(2*3*...*k)}\right]$

To have an integer it has to be that for some integer M:

$\left[\frac{(3*4*...*k) + (2*4*...*k) + ... + (2*3*...*k-1)}{(2*3*...*k)}\right] = M$

$(3*4*...*k) + (2*4*...*k) + ... + (2*3*...*k-1) = (2*3*...*k)*M$

This can be rewritten as:

$k*((k-1)*((k-2)*...(2+3) + 2*3) + 2*3*4)...) + (2*3*...*k-1) = (2*3*...*k)*M$

We can say that:

$((k-1)*((k-2)*...(2+3) + 2*3) + 2*3*4)...) = C$

So that:

$k*C + (2*3*...*(k-1)) = (2*3*...*k)*M$

We can also say that:

$(2*3*...*(k-1)) = N$

So that our final equation is:

$kC + N = (2*3*...*k)M$

The series is known as the Harmonic series, and is often dealt with for primes.

If k is prime, this is the sum of a multiple and a non-multiple of k.

Clearly there is no M that satisfies the equation.

The Harmonic Series does not give an integer for any $k\geq2$

$Q.E.D.$