Question

converge or diverge
{sum_{1}^{infty} 3^{-2} e ^{k}}

Answer 1

explain

Answer 2

\[ {\sum_{k=1}^{\infty} 3^{-2} e ^{k}} \]
Looks much better,
use similar test as before ... it's quite easier that the last one.

Answer 3

can i pull the constant out

Answer 4

yes

Answer 5

series diverges because limit approaches infinity

Answer 6

this seems too obvious, makes me suspicious something is missing

Answer 7

i thought there was more to it as well but idk

Answer 8

looks like you are missing some part of question.

Answer 9

nope thats the question

Answer 10

it's just crazy obvious with a positive exponent that\[\frac19\sum_{k=1}^{\infty}e^k\]diverges
I would think they would maybe ask you about\[\frac19\sum_{k=1}^{\infty}e^{-k}\]

Answer 11

it clearly looks divergent.

Answer 12

thanks guys