Question

I'm trying to find the sum from n=5 to INF of (10/11)^n. It seems like it ought to be easy enough, but I think the starting point of n=5 is what's throwing me off. Thanks in advance for any hints. :)

Answer 1

here is sum of a geometric series
\[\sum_{}^{} r^n = \frac{a_1 (1-r^n)}{1-r}\]
a_1 = (10/11)^5

Answer 2

note r^n goes to 0 as n->inf

Answer 3

So it's because r < 1, r^n -> 0? That's because we're raising a fraction to a higher and higher power?

Answer 4

correct, if r>1 then the infinite series would not coverge

Answer 5

Ah, I wish *I* could have converged on that idea faster. :)

Answer 6

:)

Answer 7

That was it, thanks a ton!