Question

Sequence and series..

Answer 1

\[\sum_{n \rightarrow1}^{\infty}( \frac{ (x^2-1) }{ 2 })^n\]

Answer 2

find interval of convergence>>>
Ratio and sum of infinite

Answer 3

first of all.. i use the root test to get it.. but how to get the interval.???

Answer 4

ratio test might work well

Answer 5

There is also the test of \(a_n ->0\) as \(n->\infty \)

Answer 6

impose \[\lim \frac{a_{n+1}}{a_n}<1\] then solve \(x\)

Answer 7

you get
\[\frac{a_{n+1}}{a_n}=\frac{x^2-1}{2}\] so you need
\[|\frac{x^2-1}{2}|<1\] or \[|x^2-1|<2\]

Answer 8

i got
\[-\sqrt{3}<x<\sqrt{3}\]

Answer 9

careful here. you have to solve
\[-2<x^2-1\] and
\[x^2-1<2\]

Answer 10

so
-1<x^2<3

Answer 11

oh doh you are right, ignore me

Answer 12

\[-2<x^2-1\] is always true

Answer 13

so you just need
\[x^2-1<2\] and your answer
\[-\sqrt3<x<\sqrt3\] is correct

Answer 14

so how to get ration and sum of infinite..
it is geometric series

Answer 15

*ratio

Answer 16

@satellite73
@sirm3d

Answer 17

R1=\[\frac{ x^2 -1 }{ 2}\]
and

S\[S \infty= \frac{ 2 }{ 3-x^2 }\]

Answer 18

it is correct..

Answer 19

i guess so
it is
\[\frac{a}{1-r}\] i think or else
\[\frac{1}{1-r}\] depending if you are starting at \(n=0\) or \(n=1\)

Answer 20

if start n=0 the value or R still same right.

Answer 21

yes

Answer 22

just adds a one at the beginning
i get
\[\frac{2}{3-x^2}\] if you start at \(n=0\) that is, if you compute
\[\frac{1}{1-\frac{x^2-1}{2}}\]

Answer 23

this is a geometric series, condition for convergence of geometric series might give you results.