**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.

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