Question

Evaluate sum_{5}^{infty}(\frac{ 8 }{ 9 })^n

Answer 1

@amistre64 help

Answer 2

this related to geometric series topic, yes ?
heard about it ?

Answer 3

no,haven't heard about it

Answer 4

so summation,
solved any summation formula before ?
\(\sum \limits_1^{\infty }a^n = \dfrac{a}{1-a} \)
know this formula ?

Answer 5

i got 72/9

Answer 6

did you just use that formula ?
notice that the lower limit there is +1 but the lower limit in your question is +5

Answer 7

so is it a/5-a ?

Answer 8

?

Answer 9

so, you cannot use that formula directly
you need to make a substitution
to make the lower limit from 5 to 1
you need to put k = n-4
so that when n = 5, k =1

Answer 10

when n= infinity, k= infinity
so you get
\(\sum \limits_1^{\infty }(\dfrac{8}{9})^{k+4}\)
got this ? (as k= n-4, n = k+4)

Answer 11

any doubts till here ?

Answer 12

so then i get \[\sum_{1}^{5}(\frac{ 8 }{ 9 })^5\]

Answer 13

?

Answer 14

where did get that from ?
no.....
see we need the form a^n to use that formula , right ?

Answer 15

so we do
\((8/9)^{n+4} = (8/9)^n \times (8/9)^4\)
got this ?

Answer 16

and since the constants can be taken out of summation, we have
\(\large (8/9)^4 \times \sum \limits_1^{\infty } (8/9)^n\)
now you summation is of the standard form!, use the formula i gave you

Answer 17

getting all these steps ? ask if any doubt in any step...

Answer 18

ok now what?

Answer 19

use the formula for the sum part
(8/9)^4 * (....?... )

Answer 20

sorry i dont get this .....i just started learning calculus .... could you explain to me step by step

Answer 21

ok, till which step did u clearly get ?

Answer 22

none

Answer 23

did u get why i put k = n-4
?

Answer 24

i wanted to make the lower limit = 1 to be able to use that general formula
it was 5 , so i had to subtract 4
thats why i put k = n-4
ok?

Answer 25

ok

Answer 26

so thats a standard it must always be 1 ?

Answer 27

to use that standard formula, yes
so, you got till here, right ?
\(\sum \limits_1^{\infty }(\dfrac{8}{9})^{k+4}\)

Answer 28

yes

Answer 29

the formula also required to have the exponent as just 'n' or 'k'
but we have k+4 here
so i used the exponent rule
\(x^{m+n}=x^m\times x^n\)

Answer 30

hence i got
\(\large (8/9)^4 \times \sum \limits_1^{\infty } (8/9)^n\)
as (8/9)^4 is constant and can be factored out

Answer 31

ooo, i see it now

Answer 32

the summation that we have now is of the standard form a^n
here, a = 8/9

Answer 33

so use that standard formula to find the value of summation ?

Answer 34

a/(1-a)
with a = 8/9

Answer 35

once you findd the summation value,
just multiply it with (8/9)^4
to get the final answer