summation of n^2k. k = 1 to infinity

[KataKoniK]

Is the summation of k = 1 to infinity for n2k equal to
n2 / (1 - n2)?

[amcavoy]

Is the summation of k = 1 to infinity for n2k equal to
n2 / (1 - n2)?
Yes it is (for 0<=n<1).

Alex

[KataKoniK]

Thanks for the confirmation.

[HallsofIvy]

Indeed, that is just a geometric sequence with common ratio n2.

(I will confess that my first thought was that n must mean a positive integer, for which, of course, this does not converge!)