124 WIP

n=1,2,3,4...
a_n=\left(\frac{n}{n+1}\right)^{1-n}
n=1
=\left(\frac{1}{1+1}\right)1-1=1
n=2
=\left(\frac{2}{2+1}\right)^{1-2}=\frac{3}{2}
n=3
=\left(\frac{3}{3+1}\right)^{1-3}=\frac{16}{4}=4
n=4
=\left(\frac{4}{4+1}\right)^{1-4}=\frac{125}{64}