Check the solution to "Determine the sum of the following series. ∑n=1∞(1n+9n11n)"

philosphy111of

Answered question

2021-11-13

Determine the sum of the following series.

\sum_{n = 1}^{\infty} (\frac{1^{n} + 9^{n}}{11^{n}})

Jennifer Hill

Beginner2021-11-14Added 10 answers

We have to find sum of series.
Series is given as:

\Rightarrow \sum_{n = 1}^{\infty} (\frac{1^{n} + 9^{n}}{11^{n}})

We will use geometric series test to find sum of series .
Geometric series test is given below:

\Rightarrow \sum_{n = 1}^{\infty} a r^{n - 1}

\Rightarrow

if |r|<1 (series converges)

\Rightarrow \sum = \frac{a}{1 - r}

(infinite gp)
With the help of geometric series we will find sum .
Work is shown below:

\Rightarrow \sum_{n = 1}^{\infty} (\frac{1^{n} + 9^{n}}{11^{n}})

\Rightarrow \sum_{n = 1}^{\infty} {(\frac{1}{11})}^{n} + {(\frac{9}{11})}^{n}

\Rightarrow \sum_{n = 1}^{\infty} {(\frac{1}{11})}^{n} = \frac{\frac{1}{11}}{1 - \frac{1}{11}}

\Rightarrow \sum_{n = 1}^{\infty} {(\frac{1}{11})}^{n} = \frac{1}{10}

\Rightarrow \sum_{n = 1}^{\infty} {(\frac{9}{11})}^{n} = \frac{\frac{9}{11}}{1 - \frac{9}{11}}

\Rightarrow \sum_{n = 1}^{\infty} {(\frac{9}{11})}^{n} = \frac{9}{2}

\Rightarrow \sum_{n = 1}^{\infty} {(\frac{1}{11})}^{n} + {(\frac{9}{11})}^{n} = \frac{1}{10} + \frac{9}{2}

\Rightarrow \sum_{n = 1}^{\infty} {(\frac{1}{11})}^{n} + {(\frac{9}{11})}^{n} = \frac{23}{5}

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