∏n=232n32≈2.14×10-13product from n equals 2 to 32 of n over 32 end-fraction is approximately equal to 2.14 cross 10 to the negative 13 power 1. Identify product sequence
P=2×3×4×…×323231cap P equals the fraction with numerator 2 cross 3 cross 4 cross … cross 32 and denominator 32 to the 31st power end-fraction
32!3231the fraction with numerator 32 exclamation mark and denominator 32 to the 31st power end-fraction , which is approximately (2/32)(3/32)(4/32)(5/32)(6/32)(7/32)(8/32)(9/32...
To "prepare paper" for the expression , we must first define the product's range and then calculate its value. Assuming the sequence continues until the numerator reaches the denominator's value ( ), the product is:
The following graph shows how the cumulative product decreases as more terms are added to the sequence. The product of the sequence is exactly ∏n=232n32≈2
We can rewrite the product of these 31 fractions as a single expression using factorials:
Notice that the numerator is the factorial of 32, but missing the first term ( The product of the sequence is exactly We
P=32!3231cap P equals the fraction with numerator 32 exclamation mark and denominator 32 to the 31st power end-fraction 3. Calculate the value Using the values for 323132 to the 31st power