Question

what do i compare this series to?

Answer 1

\[\sum_{n=1}^{\infty} \frac{ e ^\frac{ 1 }{ n } }{ n^2 }\]

Answer 2

@perl @SithsAndGiggles @freckles

Answer 3

i am using the comparison test and im not sure what to compare this one with

Answer 4

hello

Answer 5

isnt e^(-n) less than 1 for n>1 ?

if we compare with 1/n^2 wouldnt that suffice?

Answer 6

Is a power squared going to increase faster or is a number ^1/n going to increase faster

Answer 7

yeah i suppose we could use 1/n^2

Answer 8

spoe e^(-n) = 1/k for some n

1/k
----
n^2


1
---
kn^2

Answer 9

could you explain that again please

Answer 10

like why do we have 1/k

Answer 11

1/k is simpler to see than e^(-n) its just some fraction or approximation of a fraction

Answer 12

not a big deal but when you talk about e^(-n)
do you mean e^(1/n)

Answer 13

and I say not a big deal because both of them I think is less than 1 for n>1

Answer 14

\[\frac{1}{kn^2}<\frac{1}{n^2}\]
\[\frac{1}{k}<1\]
if 1/n^2 converges, then 1/(kn^2) converges by default since it is trapped under 1/n^2

Answer 15

what? im i reading things off ...

Answer 16

i dont think so

Answer 17

1/n = n^(-1)

e^(1/n) = e^(n^(-1))

Answer 18

do exponent rules say that can be 1/e^(n) ?

Answer 19

so an= 1/kn^2 and bn = 1/n^2

Answer 20

\[\text{ say } n=\frac{1}{2} \\ e^{\frac{1}{n}} \text{ replace } n \text{ with } \frac{1}{2} \text{ we have } e^2 \\ \text{ but } e^{-n}=e^{-\frac{1}{2}} \\ e^2>e^\frac{-1}{2}\]

Answer 21

so exp(1/n) doesn't equal exp(-n)

Answer 22

i see

Answer 23

yeah, i was looking at it a bit off :/

Answer 24

do we need to do comparison?

Answer 25

yeah