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