It turns out that what my physicist friend really wanted was the following k-fold nested summation, where each upper limit is infinite and each lower limit the next outer summation variable: Given integers n > 0, k > 0, show \sum_{i_k = 0}^oo (1/i_k) ... ... \sum_{i_2 = i_3}^oo (1/i_2) \sum_{i_1 = i_2}^oo (1/i_1) (n+1)! (i_1)! / (n+1+i_1)! = (n+1)/n^k . I know of no way even to present the left-hand side to a CAS, let alone persuade the beast to deliver the correct answer: the employment of a recursive function appears unavoidable, then there are tricky issues of delayed evaluation involved. Anyone care to comment? Fred Lunnon On Thu, Aug 12, 2010 at 6:11 AM, Fred lunnon <fred.lunnon@gmail.com> wrote:

I illustrate with the tetranomial case --- the generalisation to any dimension is obvious:

\sum_{0<=j<oo} p!q!r!j!/(p+q+r+j)! = p!q!r!/(p+q+r-1)(p+q+r-1)!

Does anyone know of a reference for these and similar results --- for example, generalisation to summation over several indices? Can Mathematica, Macsyma etc. deduce them --- Maple 9 flunked!

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

On 8/30/10, Fred lunnon <fred.lunnon@gmail.com> wrote:

I looked at trying to remove this restriction on n, which led to the slightly cleaner version attached below; but in the end it proves impossible to relax the constraints nontrivially. WFL

Given integers n >= 0, k >= 0, show

\sum_{i_k = 1}^oo (n/i_k) ...

... \sum_{i_2 = 1+i_3}^oo (n/i_2)

\sum_{i_1 = 1+i_2}^oo (n/i_1)

n! (i_1)! / (n + i_1)!

= 1 .

Silly boy --- of course it doesn't work for n = 0 --- the LHS has to be zero! But to recover my dignity, I'd just like to point out that it does apparently work for (all) _real_ n > 0, not just integers. In the code below, supersede definition of bin() by # Binomial coefficient n_C_m (via Gamma function) n >= m >= 0 only bin := proc(n, m) GAMMA(1+n)/GAMMA(1+m)/GAMMA(1+n-m) end; and "n := 10" by e.g. "n := 0.1" WFL

#### Maple code for Chris Coath summation problem (recast) ####

# For all integer n, and non-negative q,k --- # Define # g_k(q) = {n, q} when k = 0 , # = \sum_{j=q+1}^{i=\infty} g_{l-1}(j) n/j when k > 1 ; # Then # g_k(q) = {n, q} for k >= 0 .

# Binomial coefficient n_C_m (via subfactorial) fails in sum() bin := proc (n, m) local i, ans; if m < 0 then ans := 0 else ans := 1; i := 1; while i <= m do ans := ans*(n-i+1)/i; i := i+1 od fi; ans end;

# Binomial coefficient n_C_m (via library) slow bin := proc (n, m) binomial(n, m) end;

# Binomial coefficient n_C_m (via factorial) n >= m >= 0 only bin := proc(n, m) factorial(n)/factorial(m)/factorial(n-m) end;

# Definition gkD := proc(q, n, k) local j;

if k = 0 then 1/bin(n+q, q) else evalf(sum('gkD(j, n, k-1)*n/j', 'j' = q+1..infinity)) fi end;

# Inductive step gkI := proc(q, n, k) local j;

evalf(sum('gkE(j, n, k-1)*n/j', 'j' = q+1..infinity)) end; # Explicit expression gkE := proc(q, n, k); 1/bin(n+q, q) end;

# Compare inductive step with explicit expression --- approx zeros m := 6;

seq(seq(gkI(0, n, k) - gkE(0, n, k), n = 1..m), k = 0..m);

# Compare definition with explicit expression --- approx zeros (1050 sec) n := 10; m := 3; seq(gkD(0, n, k) - gkE(0, n, k), k = 0..m);