Question

What is the sum when the upper bound of summation is infinite and n=1: (1-2)^n -10

*This is sigma by the way*

so it's:
∞
Sum((1-2)^n - 10
n=1

Answer 1

Is your(1-2)^2 supposed to be (1/2)^n ? If not, it is (-1)^n, and the series will not converge.

Why do you say n=1 in your first sentence and then talk about a summation of n from 1 to infinity later?

I am assuming that the "-10" term appears only once, AFTER the series sun, and not with each term of the series. In that case, what you have written equals -9.

The limit of the sum of (1/2)^n as n goes from 1 to infinity is 2-1 = 1

Answer 2

o yes sorry, it is (1/2)^n, so:
∞
Sum((1/2)^n - 10
n=1

Answer 3

I guess that i am saying that n starts at 1 and n goes up to infinite, so how do u put the sum of that?

Answer 4

As you have been told, the sum of one-half to the nth power from n equals 1 to infinity is 1. Now subtract ten.