Question

Ofcourse this equation converges but how can I prove it?

Answer 1

u meant series?

Answer 2

\[\frac{ 3n^2 }{ 2^{n-1} }\]

Answer 3

yes

Answer 4

take ratio test

Answer 5

yes i think that will work

Answer 6

or define a function for |x|<1like this
\[ f(x) = \sum_{n=0}^\infty x^n = \frac{1}{1-x}\]

differentiate it twice and manipulate it around

Answer 7

i got the limit is (1/2) with the ratio test so it must converges

Answer 8

\[\lim_{n \rightarrow ∞}\frac{ a_{n+1} }{ a _{n} }\]???

Answer 9

yes that one

Answer 10

i dont understand the method that u r writing @experimentX

Answer 11

me neither...

Answer 12

so did u good the limit?

Answer 13

get*

Answer 14

I is in process.. I can on me computer limit with the radio test i 1/2 so it converges. But I have to do it by hand.

Answer 15

ok if u cant get it let me know

Answer 16

@julian25
How do I process
\[\frac{\frac{ 3(n+1)^2 }{ 2^n }}{ \frac{ 3n^2 }{ 2^{n-1} }}\]