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

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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Jennifer Hill
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:
$$\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}^{{n}}+{9}^{{n}}}}{{{11}^{{n}}}}}\right)}$$
$$\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{11}}}}\right)}^{{n}}+{\left({\frac{{{9}}}{{{11}}}}\right)}^{{n}}$$
$$\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{11}}}}\right)}^{{n}}={\frac{{{\frac{{{1}}}{{{11}}}}}}{{{1}-{\frac{{{1}}}{{{11}}}}}}}$$
$$\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{11}}}}\right)}^{{n}}={\frac{{{1}}}{{{10}}}}$$
$$\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{9}}}{{{11}}}}\right)}^{{n}}={\frac{{{\frac{{{9}}}{{{11}}}}}}{{{1}-{\frac{{{9}}}{{{11}}}}}}}$$
$$\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{9}}}{{{11}}}}\right)}^{{n}}={\frac{{{9}}}{{{2}}}}$$
$$\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{11}}}}\right)}^{{n}}+{\left({\frac{{{9}}}{{{11}}}}\right)}^{{n}}={\frac{{{1}}}{{{10}}}}+{\frac{{{9}}}{{{2}}}}$$
$$\displaystyle\Rightarrow{\sum_{{{n}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{11}}}}\right)}^{{n}}+{\left({\frac{{{9}}}{{{11}}}}\right)}^{{n}}={\frac{{{23}}}{{{5}}}}$$