Finding the sum of an infinite series with (i/2)^(2n)

[szklany]

Find the sum of the infinite series

$$\sum _{n=1}^{\infty } \left( i /2\right) ^{2\,n}$$

I just can't seem to get started on this problem, so I was hoping somebody could give me a hint, as to what methods i should read up on.

 

[clamtrox]

A good place to start would be to consider what (i/2)^2 is, that simplifies the sum considerably. You also need the formula for geometric series, that you can find for example from wikipedia.

 

[r.a.c.]

Actually, break i into its polar form which is $$i^2^n = e^i^n^(^p^i^)$$ . Now expand using Euler's identity and you are left only with $$cos\ n(pi) = (-1)^n$$ From there it is a geometric series.

 

[HallsofIvy]

Actually, there is no need to worry about "i". $(i/2)^{2n}= (i^2/4)^n= (-1/4)^n$ so this is a purely real geometric series.