Question

Find the sum of the following infinite geometric series

Answer 1

Answer:

$\large \boxed{\ \ \dfrac{63}{5} \ \ }$

Step-by-step 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

    $a_1=\dfrac{4}{9}\cdot a_0=7*\boxed{\dfrac{4}{9}}=\dfrac{7*4}{9}=\dfrac{28}{9}$

Then

   $a_2=\dfrac{4}{9}\cdot a_1=\dfrac{28}{9}*\boxed{\dfrac{4}{9}}=\dfrac{28*4}{9*9}=\dfrac{112}{81}$

and...

   $a_3=\dfrac{4}{9}\cdot a_2=\dfrac{112}{81}*\boxed{\dfrac{4}{9}}=\dfrac{112*4}{9*81}=\dfrac{448}{729}$

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

   $a_k=a_0\cdot (\dfrac{4}{9})^k$

So, for n positive integer

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

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

   $\large \boxed{\ \ \dfrac{63}{5} \ \ }$

as

   $\dfrac{4}{9}<1$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you