Rats --- goofed again --- thanks for spotting that howler, Edwin. The lower limits should all have been incremented by unity. I'm actually very pleased, because in correcting it I saw how to get rid of that pesky factor (n+1) which had been hanging around and making the place look untidy. So the corrected version now reads (note k-fold nested summation): Given integers n > 0, k > 0, show \sum_{i_k = 1}^oo (1/i_k) ... ... \sum_{i_2 = 1+i_3}^oo (1/i_2) \sum_{i_1 = 1+i_2}^oo (1/i_1) n! (i_1)! / (n + i_1)! = 1 / n^k . Perhaps I should make it clear that I have already solved the problem. What I'm interested in here is whether it is possible to pose it formally to a CAS at all, rather than whether a CAS can solve it. Fred Lunnon #### Maple code for Chris Coath summation problem #### # Define # h_k(q) \equiv \sum_{j=q+1}^{j=\infty} {n, j}/j for k = 1 , # \sum_{j=q+1}^{i=\infty} h_{l-1}(j)/j for k > 1 ; # Then # h_k(q) = 1/n^k {n, q} . # Reciprocal multinomial coefficient {p_1,...,p_k} RMC := proc(plis) local i; product(factorial(plis[i]), i = 1..nops(plis)) / factorial(add(plis[i], i = 1..nops(plis))) end; # Definition hkD := proc(q, n, k) local j; if k = 1 then evalf(sum('RMC([n, j])/j', 'j' = q+1..infinity)) else evalf(sum('hkD(j, n, k-1)/j', 'j' = q+1..infinity)) fi end; # Inductive step hkI := proc(q, n, k) local j; evalf(sum('hkE(j, n, k-1)/j', 'j' = q+1..infinity)) end; # Explicit expression hkE := proc(q, n, k); 1/n^k*RMC([n, q]) end; # Compare inductive step with explicit expression --- approx zeros m := 6; seq(seq(hkI(0, n, k) - hkE(0, n, k), n = 1..m), k = 1..m); # Compare definition against explicit expression --- approx zeros (slow!) n := 10; m := 3; seq(hkD(0, n, k) - hkE(0, n, k), k = 1..m); On 8/29/10, W. Edwin Clark <wclark@mail.usf.edu> wrote:
Fred,
I am not certain what the summation is. For one thing i_k = 0 is not possible for 1/i_k. I tried feeding it to Maple for specific values of n and k = 1 (and k = 2), with i_k = 1 to infinity, unsuccessfully. But I'm not sure I understand the formulas.
Can you send Maple code for cases k = 1, 2 and 3? Using assume(n::posint), Maple might be able to verify for variable n. But at least one can check by putting in values for n.
Edwin