Question

to where does the sum of the series e^(-3n) converge? n lies upon [1, infinity]

Answer 1

@AriPotta @Hope_nicole @SnuggieLad any suggestions or who could help?

Answer 2

O_O

Answer 3

sorry man...

Answer 4

well i think i know how to graph it..

Answer 5

\[\large e^{-3n}=(e^{-3})^n\]
you have a geometric series

Answer 6

@sirm3d can you help me solve this problem? I haven't done many problems with convergences not at 0

Answer 7

what do you know about geometric series?

Answer 8

if abs(r) is less than one the series will converge?

Answer 9

but I don't know how to solve for the point of convergence

Answer 10

okay.

what is \( r\) in the series in question?

Answer 11

1/e^3

Answer 12

a/(1-r)

Answer 13

right. r=1/e^3 and the first term a is also 1/e^3

just sub them in into the sum \[\frac{a}{1-r}\]