nevermind, yes, it is a slow morning...day.

Find Limit[1/n^2*Sum[Sum[Sum[1/(i+j+k),{i,1,n}],{j,1,n}],{k,1,n}],n->Infinity]HintLimit[1/n*Sum[Sum[1/(i+j),{i,1,n}],{j,1,n}],n->Infinity] = 2*Log[2]

The limit equals the integral of 1/(x+y+z) over the cube 0< x,y,z<1. Use iterated integration and the usual primitives for the functions of the form u^n log(u) (e.g. log(u) --> u log(u)-u etc)....i ran out of napkins....

replace sums by integrals with an error that probably goes to zero after you divide by n^2integrate by parts a couple times to deal with the ln(x) terms until you've surely made an algebra mistakeand with a little luck you get the answer which is probably an integer times log(3)

The multiple of log 3 should be obtained from going from1/(x+y+z) to log(x+y+1) to (x+2) log(x+2) to u^2/2 log(u) where u=3so it should be 9/2

ok, I propose(9/2) log(3) -3 +2 log(2)

you definitely have more napkins than I do

I work in a bar at night....you know....to make ends meet.... After a few drinks, things are getting clearer and +s have turned into -s giving the limit as:(9/2)log(3)-6log(2)

you mean, make the 'late' end of today meet the 'early' end of tomorrow? that's my bar strategy anyhow

So, it looks like the limit is smaller for N=3 in comparison with the limit for N=2.Here N is a number of variables and summations in the formula Limit[1/n^(N-1)*Sum...Sum(1/(i+j+k+...)), n->Infinity].N=2 1.386... = 2*Log[2]N=30.785... = - 6*Log[2] + 9/2*Log[3]N=40.557... = 88/3*Log[2] - 18*Log[3]N=50.433... = - 5/24*(544*Log[2] - 162*Log[3] - 125*Log[5])It looks like the results depend on N.Is there an asymptotic formula for such Limits when number of variables goes to Infinity? (N->Infinity)

well, the problem is that the first term is 1/N... if you had put index sums >=0 rather than >=1, and added a 1 to the denominator, the problem would be much easier to solveI didn't do it, but am pretty certain that the lim is 0, probably something on the order of log N / N

chukchi, are u sure in coeffs for N=4? i generate similar expansions into logs of prime numbers, but for N=4 i have a different coeffs.

This is a generalization of the well known Euler's function:I(n,N) = Integrate[(x*(1-x^n)/(1-x))^N/x,x]/n^(N-1)N = 4Evaluate at x=1Limit I(n,4), n->InfinityorIntegrate[1/(x+y+z+t),{x,0,1},{y,0,1},{z,0,1},{t,0,1}] = 88/3*Log[2] - 18*Log[3] = 0.557296100...

The question is "What is the limit L(N) = I(Infinity,N) as N->Infinity?"

QuoteOriginally posted by: krwell, the problem is that the first term is 1/N... if you had put index sums >=0 rather than >=1, and added a 1 to the denominator, the problem would be much easier to solveI didn't do it, but am pretty certain that the lim is 0, probably something on the order of log N / NActually, the 2-dimensional problem for Sum of matrix elements 1/(i+j) is related to, so called, Hilbert Matrix.Our matrix is identical to Hilbert Matrix without the first row and the last column. The difference in Sums of all matrix elements for our matrix and the Hilbert Matrix is equal to Harmonic Number (n = 2N - 1).Our matrix could be considered as a generalization of Harmonic Number into 2 dimensions. It is interesting to find a simple expression of our Sums via Harmonic Numbers in 2 and higher dimensions.