Question #f639b
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"How do I generate the electron configuration of iron?"
Use the Ratio Test (it's usually a good idea whenever factorials are involved) to show that this series converges.
Let #a_{n}=(2/n)^{n}*n!#.
We want to know whether the series #sum_{n=1}^{infty}a_{n}# converges or not.
To apply the Ratio Test, we consider the sequence #|a_{n+1}/a_{n}|#. In this example, all the terms in the original sum are positive, so we may dispense with the absolute value signs and note that #a_{n+1}/a_{n}=(2/(n+1))^(n+1) * (n+1)!*(n/2)^(n)*1/(n!)#.
This expression simplifies to:
#a_{n+1}/a_{n}=2(n/(n+1))^(n)#.
It is well known that #((n+1)/n)^{n}=(1+1/n)^{n}->e# as #n-> infty#. Therefore, #a_{n+1}/a_{n} -> 2/e approx 0.73576<1# as #n-> infty#.
By the Ratio Test, this is enough to imply that the series #sum_{n=1}^{infty}(2/n)^{n}*n!# converges.
Using the Stirling asymptotic formula
#n! approx sqrt(2pin)(n/e)^n# we have
#sum_(n=1)^oo(2/n)^2n! approx sum_(n=1)^oo(2/n)^2sqrt(2pin)(n/e)^n = sum_(n=1)^oo sqrt(2pin)(2/e)^n# and
#sum_(n=1)^oo sqrt(2pin)(2/e)^n le sum_(n=1)^oo (n+1)(2/e)^n#
and from
#sum (k+1)x^k = d/dx(sum x^(k+1))# the convergence for #sum (k+1)x^k# is assured if #abs x < 1# or in our case
#2/e < 1# and as a consequence
#sum_(n=1)^oo(2/n)^2n! # converges.
