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Dominik [7]
3 years ago
5

Identify whether the series summation of 8 open parentheses 5 over 6 close parentheses to the i minus 1 power from 1 to infinity

is a convergent or divergent geometric series and find the sum, if possible. (2 points)
Mathematics
2 answers:
musickatia [10]3 years ago
5 0
<h2>Answer:</h2>

Yes, the series is convergent.

The sum is: 48

<h2>Step-by-step explanation:</h2>

The series is given by:

\sum_{i=1}^{\infty} 8(\dfrac{5}{6})^{i-1}

Now, each term of the series is a constant multiple of the preceding element of the series.

The constant multiple is: 5/6<1

( Since,

a_1=8(\dfrac{5}{6})^{1-1}\\\\i.e.\\\\a_1=8(\dfrac{5}{6})^0\\\\i.e.\\\\a_1=8

a_2=8(\dfrac{5}{6})^{2-1}\\\\i.e.\\\\a_2=8(\dfrac{5}{6})\\\\i.e.\\\\a_2=\dfrac{5}{6}a_1

Similarly, for nth term we have:

a_n=\dfrac{5}{6}a_{n-1}

)

Hence, the series is a geometric series.

Also, this series is convergent.

( since the constant multiple i.e. the common ratio is less than 1)

We know that the sum of the infinite geometric series of the type:

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

where a is the first term of the series and r is the common ratio.

Here we have:

a=8\ and\ r=\dfrac{5}{6}

Hence, we have:

\sum_{i=1}^{\infty} 8(\dfrac{5}{6})^{i-1}=\dfrac{8}{1-\dfrac{5}{6}}\\\\i.e.\\\\\sum_{i=1}^{\infty} 8(\dfrac{5}{6})^{i-1}=\dfrac{8}{\dfrac{6-5}{6}}\\\\i.e.\\\\\sum_{i=1}^{\infty} 8(\dfrac{5}{6})^{i-1}=\dfrac{8}{\dfrac{1}{6}}\\\\i.e.\\\\\sum_{i=1}^{\infty} 8(\dfrac{5}{6})^{i-1}=8\times 6\\\\i.e.\\\\\sum_{i=1}^{\infty} 8(\dfrac{5}{6})^{i-1}=48      

Leto [7]3 years ago
3 0
\displaystyle\sum_{i=1}^\infty\8\left(\dfrac56\right)^{i-1}

This is a geometric series with a common ratio between successive terms that is less than 1 in absolute value. This means the series will converge. The value of the sum is \dfrac8{1-\frac56}=48.

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