Question

How to get it?
\[\large \sum\limits_{k\geq n} \dfrac{1}{2^k}= \dfrac{1}{2^{n-1}}\]
Please help

Answer 1

can type it up again plzzz

Answer 2

its not readable

Answer 3

I read but don't know how to get this part.

Answer 4

isnt it just a geometric series ?

Answer 5

yes, but we have formula to get it as = \(a+ar+ar^2+.....+ar^n =a\dfrac{1-r^{n+1}}{1-r}\) for 0 <r <1

Answer 6

\[\large S_{\infty} = \dfrac{a}{1-r}\]

Answer 7

infinite series formula requires |r| < 1
but your partial sum formula doesn't require it

Answer 8

the last term is ar ^(n-1)

Answer 9

there is no last term, its an infinite series
k >= n means, k = m to infinity

Answer 10

yes, I understand it

Answer 11

but ganesh i have little doubt dont mind

Answer 12

here our first term will be 2^n right not 1???

Answer 13

yes a = 1/2^n
r = 1/2

right ?

Answer 14

yes

Answer 15

yeah i mean that...okkkk fine :)

Answer 16

just conforming ..that we r goin to start with n :P

Answer 17

But in this stuff, it doesn't last long to infinitive, it starts from m and stop at m+1
then take sum of those part give us ... something like above, but I don't get how to get 1/2^(n-1)

Answer 18

n will be a real number

Answer 19

but upper limit to k is not given

Answer 20

any partial sum is less than the infinite sum because the series is increasing

Answer 21

Got you!! make perfect sense to me to skip the part I don't get.

Answer 22

\[\large \sum\limits_{k\ge m} \dfrac{1}{2^k} ~\gt ~~ \sum\limits_{k= m}^{m+k-1} \dfrac{1}{2^k} \]

Answer 23

next?

Answer 24

they are using that

Answer 25

In the attached doc :

the partial sum in line3 is less than tha infinite sum in line4

Answer 26

its like saying this :

10^2 = 10^2
< 10^3
< 10^4
= 10000

Answer 27

I have class now, I have to go. Thanks for the help

Answer 28

the end result is that

10^2 < 10000

Answer 29

Okay, its simple for you if u look at it calmly :)

Answer 30

Again, thanks a lot