Question

How do you find the sum of the sum of the infinite series \[\sum {{{\left( {\dfrac{1}{2}} \right)}^n}\,from\,n = 0\,to\,\infty } \] ?

Answer 1

Hint:The given question is needed to be solved by expanding the term in additional form and then add every term and get the final solution for the question. Here we can also use the formula for the infinite summation of half starting from one to infinity then will that answer we can add the value of half for the power zero.

Formulae Used:
\[{S_n} = 1 - \dfrac{1}{{{2^n}}}\], for summation of an infinite series of half, from one to infinity.

Complete step by step answer:
The given question is \[\sum {{{\left( {\dfrac{1}{2}} \right)}^n}\,from\,n = 0\,to\,\infty } \]
Here we will break the question into two parts that is we know the summation for the series from one to infinity, and for the power zero we will add the answer to the series, after breaking these series we get:
\[\sum {{{\left( {\dfrac{1}{2}} \right)}^n}\,from\,n = 0\,to\,\infty } \\
\Rightarrow \sum {{{\left( {\dfrac{1}{2}} \right)}^0} + \sum\limits_1^\infty {{{\left( {\dfrac{1}{2}} \right)}^n}} } \\
\Rightarrow 1 + \left( {\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + .... + \dfrac{1}{{{2^{n - 1}}}} + \dfrac{1}{{{2^n}}}} \right) \\ \]
Now we have to solve the second bracket for which we already know the formulae which is:
\[\sum\limits_1^\infty {\left( {\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + .... + \dfrac{1}{{{2^{n - 1}}}} + \dfrac{1}{{{2^n}}}} \right)} = 1 - \dfrac{1}{{{2^n}}}\]
Here we have to put the value of variable “n” as one and infinity, on solving we get:
\[for\,n = 0,1 - \dfrac{1}{{{2^n}}} = 0 \\
\Rightarrow for\,n = \infty ,1 - \dfrac{1}{{{2^n}}} = 1\,because\,\dfrac{1}{{{2^\infty }}} = 0 \\ \]
So now our required answer is: \[1 + 0 + 1 = 2\]

Hence, the sum of the sum of the infinite series is two.

Note:Here this question can be solved by this method only, and we need to apply the formulae of the infinite series, because if it is not done then we have to use the geometric series formula which is same what we used here, and will give more steps to solve the question, hence above is most suitable method to solve.