Mathematics

Question

identify whether the series infinity sigma i=1 8(5/6)^i-1 is a convergent or divergent geometric series and find the sum if possible

2 Answer

  • 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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