How do you find a formula for the sum n terms #Sigma (1+(2i)/n)^2(2/n)# and then find the limit as #n->oo#?

Question

5870 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-26T21:41:45

The sum diverges.

#(1+(2i)/n)^2(2/n) = (1+2i(2/n)-(2/n)^2)(2/n)=#
#2/n+2i(2/n)^2-(2/n)^3#

then

#sum_(n=1)^oo(1+(2i)/n)^2(2/n)=2sum_(n=1)^oo1/n+8isum_(n=1)^oo1/n^2-8sum_(n=1)^oo1/n^3#

Here

#sum_(n=1)^oo1/n# diverges
#sum_(n=1)^oo1/n^2=pi^2/6#
#sum_(n=1)^oo1/n^3=zeta(3)approx1.20206#

so the sum

#sum_(n=1)^oo(1+(2i)/n)^2(2/n)# diverges