Question

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)

Answer 1

Answer:

Yes, the series is convergent.

The sum is: 48

Step-by-step explanation:

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:

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

Answer 2

$\displaystyle\sum_{i=1}^\infty\8\left(\dfrac56\right)^{i-1}$

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 $\dfrac8{1-\frac56}=48$.