Question #be944
1 Answer
Dec 20, 2016
If we expand the summation as follows
#1/n sum_{k=0}^(n-1) e^{k/n} = 1/n { e^(0/n)+e^(1/n)+e^(2/n) + e^(3/n) + ... + e^((n-1)/n) } #
#" " = 1/n { underbrace(e^0+e^(1/n)+(e^(1/n))^2 + (e^(1/n))^3 + ... + (e^(1/n))^(n-1))_("n terms") } #
So you are correct, It is a GP with;
first term
#a=e^0 (=1)# and
common ratio#r=e^(1/n)#
So Using the GP summation formula:
#S_n=a((1-r^n))/((1-r))#
to get:
#1/n sum_{k=0}^(n-1) e^{k/n} = 1/n1((1-(e^(1/n))^n))/((1-e^(1/n)))#
#" " = ((1-e^(n/n)))/(n(1-e^(1/n)))#
#" " = ((e^(n/n)-1))/(n(e^(1/n)-1))#
#" " = ((e^(n/n)-e^0))/(n(e^(1/n)-1))# (as#e^0=1# ) QED