Question #db134

1 Answer
yumean1119
Nov 2, 2017

About #3.455*10^9014#

Explanation:

This can be caluculated at
http://www.wolframalpha.com/input/?i=product+of+n%5En+from+n%3D1+to+n%3D100

This type of sequence is called hyperfactorial.

https://oeis.org/A002109
http://googology.wikia.com/wiki/Hyperfactorial

[Approximation]
Let the product #P=prod_(n=1)^100n^n=1^1*2^2* * * 100^100#.
Then,
#ln P=sum_(n=1)^100nlnn# and
#int_1^100xlnxdx< ln P < int_1^101xlnxdx#
(See about Riemann sum: https://en.wikipedia.org/wiki/Riemann_sum)

Since #intxlnxdx=(x^2lnx)/2 -x^2/4 +C#,
#[(x^2lnx)/2 -x^2/4]_1^100 < ln P < [(x^2lnx)/2 -x^2/4]_1^101#
#5000ln100 - 2499.75 < lnP < 5100.5ln101 - 2550#
#20526.1 < ln P < 20989.4#
#e^20526.1< P < e^20989.4#
#2.355*10^8914 < P < 3.807*10^9115#

The geometric mean between the lower bound and the upper bound is
#sqrt(2.355*10^8914* 3.807*10^9115)=9.469*10^9014#
and this is a bit larger than the exact value of P.

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