Find the sum of the following infinite geometric series

Answer:
[tex]\large \boxed{\ \ \dfrac{63}{5} \ \ }[/tex]
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 [tex]+\infty[/tex]
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
[tex]\displaystyle \sum_{k=0}^{+\infty} a_k[/tex]
First of all, we need to find an expression for [tex]a_k[/tex]
First term is
[tex]a_0=7[/tex]
Second term is
[tex]a_1=\dfrac{4}{9}\cdot a_0=7*\boxed{\dfrac{4}{9}}=\dfrac{7*4}{9}=\dfrac{28}{9}[/tex]
Then
[tex]a_2=\dfrac{4}{9}\cdot a_1=\dfrac{28}{9}*\boxed{\dfrac{4}{9}}=\dfrac{28*4}{9*9}=\dfrac{112}{81}[/tex]
and...
[tex]a_3=\dfrac{4}{9}\cdot a_2=\dfrac{112}{81}*\boxed{\dfrac{4}{9}}=\dfrac{112*4}{9*81}=\dfrac{448}{729}[/tex]
Ok we are good, we can express any term for k integer
[tex]a_k=a_0\cdot (\dfrac{4}{9})^k[/tex]
So, for n positive integer
[tex]\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}}][/tex]
And the limit of that expression when n tends to [tex]+\infty[/tex] is
[tex]\large \boxed{\ \ \dfrac{63}{5} \ \ }[/tex]
as
[tex]\dfrac{4}{9}<1[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you