POLYA021: Adding lots of Egyptian fractions  (proposed by Len M. Smiley; 12/6/01).

For a positive integer m, calculate  lim_{n \to \infty}   sum _{1 \le i_1 \le i_2 \le ... \le i_n \le m} 1/(i_1.i_2... i_n).

Discussion. [LMS, 12/7/01]: In other words, evaluate the limit as n approaches infinity of the following sum, whose number of terms increases with increasing n.
There is one summand for each nondecreasing n-tuple  (i_1,...,i_n) satisfying 1<=i_1 and i_n<=m. This summand is the "Egyptian Fraction" with numerator 1 and denominator equal to the n-factor product i_1*...*i_n.
Note that, if m=2 , for each n we have the partial sum of the geometric series 1+1/2+1/4+...+1/2^n and so the limit is 2.

[PY, 12/7/01]: The given sum factors as the product of the m-1 infinite geometric series:
 2     = 1 + 1/2 + 1/2^2 + ... + 1/2^n + ...
3/2   = 1 + 1/3 + 1/3^2 + ... + 1/3^n + ...
4/3   = 1 + 1/4 + 1/4^2 + ... + 1/4^n + ...
...
m/(m-1) = 1 + 1/m + 1/m^2 + ... + 1/m^n + ...
From these, it follows that the sum is 2(3/2)(4/3)...(m/(m-1)) = m.
 

Bibliography.

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