Question

Find the sum of the following infinite geometric series

Answer 1

`\large \boxed{\ \ (63)/(5) \ \ }`

Explanation:

Hello,

"Find the sum of the following infinite geometric series"

infinite

We will have to find the limit of something when n tends to
`+\infty`

geometric series

This is a good clue, meaning that each term of the series follows a geometric sequence. Let's check that.

The sum is something like

`\displaystyle \sum_(k=0)^(+\infty) a_k`

First of all, we need to find an expression for
`a_k`

First term is

`a_0=7`

Second term is

`a_1=(4)/(9)\cdot a_0=7*\boxed{(4)/(9)}=(7*4)/(9)=(28)/(9)`

Then

`a_2=(4)/(9)\cdot a_1=(28)/(9)*\boxed{(4)/(9)}=(28*4)/(9*9)=(112)/(81)`

and...

`a_3=(4)/(9)\cdot a_2=(112)/(81)*\boxed{(4)/(9)}=(112*4)/(9*81)=(448)/(729)`

Ok we are good, we can express any term for k integer

`a_k=a_0\cdot ((4)/(9))^k`

So, for n positive integer

`\displaystyle \sum_(k=0)^(n) a_k=\displaystyle \sum_(k=0)^(n) 7\cdot ((4)/(9))^k=7\cdot (1-((4)/(9))^(n+1))/(1-(4)/(9))=(7*9*[1-((4)/(9))^(n+1)])/(9-4)=(63)/(5)\cdot [1-((4)/(9))^(n+1)}]`

And the limit of that expression when n tends to
`+\infty` is

`\large \boxed{\ \ (63)/(5) \ \ }`

as

`(4)/(9)<1`

Hope this helps.

Do not hesitate if you need further explanation.

Thank you