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

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

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

The sum is

[tex]\dfrac8{1-\frac56}=48[/tex]

since

[tex]\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}[/tex]

if [tex]|r|<1[/tex].

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:

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

which could also be written as:

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

As we know that any infinite series of the form:

[tex]\sum^{\infty}_{i=1}x^i[/tex]

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:

[tex]S=\dfrac{a}{1-r}[/tex]

Here we have:

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

Hence, the sum of series is:

[tex]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[/tex]

Hence, the sum of series is:

40

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