Question

sum from n=1 -> infinity of
(9^n) /(3+ 10^n)

Answer 1

\[\sum_{n=1}^{\infty} = (9^n)/ (3 +10^n)\]

Answer 2

convergent or divergent? I'm not able to use the Satellite's eyeball test for this one

Answer 3

no but almost
ignore the 3

Answer 4

the n's could be anything

Answer 5

or since you are doing mathematics and not eyeballing say \(\frac{9^n}{3+10^n}<\frac{9^n}{10^n}\)

Answer 6

the denominator will always be greater

Answer 7

and since \(\frac{9^n}{10^n}=.9^n\) you win

Answer 8

divergent?

Answer 9

oh no \(.9<1\)

Answer 10

VICTORY!!!!!

Answer 11

\[\sum_{k=1}^{\infty}(\frac{9}{10})^n=\frac{.9}{1-.9}=\frac{.9}{.1}=90\]

Answer 12

oops i mean 9

Answer 13

it's convergent because it's a geometric series

Answer 14

correct?

Answer 15

yes indeed
geometric with \(r=.9<1\)

Answer 16

great thanks!

Answer 17

yw