Determine the sum of the following series. \sum_{n=1}^\infty (\frac{1^n+9^n}{11^n})

philosphy111of 2021-11-13 Answered
Determine the sum of the following series.
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}^{{n}}+{9}^{{n}}}}{{{11}^{{n}}}}}\right)}\)

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Expert Answer

Jennifer Hill
Answered 2021-11-14 Author has 7701 answers
We have to find sum of series.
Series is given as:
\(\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}^{{n}}+{9}^{{n}}}}{{{11}^{{n}}}}}\right)}\)
We will use geometric series test to find sum of series .
Geometric series test is given below:
\(\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{a}{r}^{{{n}-{1}}}\)
\(\displaystyle\Rightarrow\) if |r|
\(\displaystyle\Rightarrow\sum={\frac{{{a}}}{{{1}-{r}}}}\) (infinite gp)
With the help of geometric series we will find sum .
Work is shown below:
\(\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}^{{n}}+{9}^{{n}}}}{{{11}^{{n}}}}}\right)}\)
\(\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{11}}}}\right)}^{{n}}+{\left({\frac{{{9}}}{{{11}}}}\right)}^{{n}}\)
\(\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{11}}}}\right)}^{{n}}={\frac{{{\frac{{{1}}}{{{11}}}}}}{{{1}-{\frac{{{1}}}{{{11}}}}}}}\)
\(\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{11}}}}\right)}^{{n}}={\frac{{{1}}}{{{10}}}}\)
\(\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{9}}}{{{11}}}}\right)}^{{n}}={\frac{{{\frac{{{9}}}{{{11}}}}}}{{{1}-{\frac{{{9}}}{{{11}}}}}}}\)
\(\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{9}}}{{{11}}}}\right)}^{{n}}={\frac{{{9}}}{{{2}}}}\)
\(\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{11}}}}\right)}^{{n}}+{\left({\frac{{{9}}}{{{11}}}}\right)}^{{n}}={\frac{{{1}}}{{{10}}}}+{\frac{{{9}}}{{{2}}}}\)
\(\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{11}}}}\right)}^{{n}}+{\left({\frac{{{9}}}{{{11}}}}\right)}^{{n}}={\frac{{{23}}}{{{5}}}}\)
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