Verified Answer :

LammettHash

The sum is convergent. I'll assume the 8 isn't an attempt at using the infinity symbol, so that you have

\(\displaystyle\sum_{i=1}^\infty 8\left(\frac56\right)^{i-1}\)

This converges because the common ratio between terms is smaller than 1.

The sum is

\(\dfrac8{1-\frac56}=48\)

since

\(\displaystyle\sum_{i=1}^\infty ar^{i-1}=a\lim_{n\to\infty}\sum_{i=1}^nr^{i-1}=a\lim_{n\to\infty}\frac{1-r^n}{1-r}=\frac a{1-r}\)

if \(|r|<1\).

Answer:

The sum of infinite geometric series is:

40

Step-by-step explanation:

We have to find the sum of the geometric series which is given as:

\(\sum^{\infty}_{i=1} 8\times (\dfrac{5}{6})^i\)

which could also be written as:

\(8\sum^{\infty}_{i=1} (\dfrac{5}{6})^i\)

As we know that any infinite series of the form:

\(\sum^{\infty}_{i=1}x^i\)

is convergent if |x|<1

Here we have:

x=5/6<1

Hence,the infinite series is convergent.

Also we know that for infinite geometric series the sum is given as:

\(S=\dfrac{a}{1-r}\)

Here we have:

a=5/6 and common ration r=5/6

Hence, the sum of series is:

\(8\sum^{\infty}_{i=1} (\dfrac{5}{6})^i=8\times (\dfrac{\dfrac{5}{6}}{1-\dfrac{5}{6}})\\\\=8\times (\dfrac{\dfrac{5}{6}}{\dfrac{1}{6}})\\\\=8\times 5\\\\=40\)

Hence, the sum of series is:

40

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