Does this help? [tex]\sum_{j = 1}^{N} \frac {j}{j + 1} = \sum_{j = 1}^{N} \left( 1 - \frac {1}{j + 1}\right) = N - \sum_{j = 1}^{N} \frac {1}{j + 1}[/tex]
Minor quibble-Tide you lost a negative sign. Tide has reduced this nicely to showing any the Harmonic series never hits an integer. This might be tough to prove so I'll kick out a hint. Let 2^k be the largest power of 2 less than or equal to N, so 2^k<=N<2^k+1. Assume the sum is an integer. Clear the denominators carefully so every term becomes an integer, except the 1/2^k term. This gives a contradiction. I'll leave it to you how to clear them 'carefully', but seeing how to do this for the first few values of N might help.
if you clear denominators you see that n! must divide the top. It follows for example that n must divide (n-1)! which is impossible say, if n is prime. If not, and p is the largest prime number less than n, it follows again that p must divide some number between p+1 and n. I.e. n must be at least 2p. But this contradicts the fact that there must be another prime between p and 2p, hence another between p and n.
This is kind of related, is: [tex]\sum_{r=1}^{n} r![/tex] ever a perfect square? I know this is true for n=1 and n=3, but what about in general? I'd like hints only, please.
I'd probably pick some useful number and look at things mod that number. for it is divisible by 3 for n>1, so it must be WHAT if it is a square?
I swear I was just thinking the same thing! m! where m>4 always ends in zero. So when n>4, we have 10p+(4!+3!+2!+1!)=10p+30+3=10q+3, but I really didn't know that if a number ends in 3 it's a perfect square.
just work out the squares of the residues mod 10, which is waht i meant by considering a suitable mod thing and if 3 divides N and N is a square then 9 divides N is what I meant
For the sake of curiosity... What you're saying is: if N>3 and N = 0 (mod 3), then N = 0 (mod 9). Right? Can we generalize this to: if N>3^n and N = 0 (mod 3), then N = 0 (mod 3^{n+1})?
This is realiant on N being a square, you realize, and nothing to do with N being greater than some power of three, nor for that matter is it to do with 3. Let N be the n'th power of q where q is an integer. Write q as the product of primes, now, what is the decomposition of N relative to q? For example, let n=3 so we#re considering cubes. What are some cubes? 8, 27, 216. How may times do 2 and 3 occur each decomposition? it's always a multiple of n isn't it?
I know this might sound naive, but why is that? Wait, nevermind... I figured it out. So, to sum up, if q divides p^n then so does q^n? Thanks for all your help matt.
Did you write out q as a product of primes? Raise it to the power n? What do you get? for instance if q=12=2^2.3, then q^3= 2^^.3^3 doesn't it?
in response to your last corrected qusetion, only if the divisor is a prime, it is one of the equivalent criteria for being prime (in the integers)
And it is not! As I had said, if a number ends in 3, it is not a perfect square. There is no n>4 which satisfies your equation.
matt, I have one more question about your proof. S = 33 (mod 10), but does S really need to be divisible by 3? 43 = 33 (mod 10), but 43 isn't divisible by 3.