How do I calculate the limit of a sum

How do I calculate the limit of this problem.

lim x-->infinity sum of (-2/3)^k (upper limit n, k=2)

I am not english and hopefully nothing is lost in the translation. thanks

This is an infinite sum of a geometric progression.
The formula for this infinite sum is S = a/(1-r) where a is the first term, and r is the common ratio.

First term = (-2/3)^2 = 4/9
common ratio = -2/3

S = (4/9) / [1 - (-2/3)]
= 4/15

S = lt n -> ∞: ∑{k =2 ..n, (-2/3)^k} = (-2/3)^2. { 1 + (-2/3) + (-2/3)^2 + ... (-2/3)^n-2 }

{ } is a GS with initial term 1 and common ratio -2/3 ==>
S(n) = 1/(1 -(-2/3)) = 1/ (1 + 2/3) = 3/5

S = 4/9 * 3/5 = 4/15