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Thread: Summation of a series

  1. #1
    Newbie AgentSmith's Avatar
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    Summation of a series

    Let $\displaystyle S_k$, $\displaystyle k = 1,\ 2, \cdots ,\ 100$, denote the sum of the infinite geometric series whose first term is $\displaystyle \frac{k-1}{k!}$ and common ration is $\displaystyle \frac{1}{k}$ then find the value of $\displaystyle \frac{100^2}{100!}+\sum_{k=1}^{100}|(k^2-3k+1)S_k|$.

    How do I do this?
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  2. #2
    Newbie AgentSmith's Avatar
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    Re: Summation of a series

    Quote Originally Posted by CaptainBlack View Post
    Have you considered starting by writing $\displaystyle S_k$ as a function of $\displaystyle k$?
    CB
    Yes. I got $\displaystyle S_k=\frac{1}{(k-1)!}$.
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  3. #3
    Member sbhatnagar's Avatar
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    Re: Summation of a series

    $\displaystyle S_k$ is an infinite geometric series.

    1.Sum of an infinite geometric series is given by $\displaystyle S=\frac{a}{1-r}$

    therefore $\displaystyle S_k= \frac{\frac{k-1}{k!}}{1-\frac{1}{k}}$

    $\displaystyle \implies S_k = \frac{1}{(k-1)!}$

    2. Note That:
    $\displaystyle \begin{align*} \sum_{k=1}^{100}\Big| \frac{k^2-3k+1}{(k-1)!}\Big|&= \sum_{k=1}^{100}\Big| \frac{(k-1)^2-k}{(k-1)!}\Big| \\ &=\sum_{k=1}^{100}\Big| \frac{k-1}{(k-2)!}-\frac{k}{(k-1)!}\Big| \\ &= |-1|+|-1|+\Big| \frac{2}{1!}-\frac{3}{2!}\Big|+\Big| \frac{3}{2!}-\frac{4}{3!}\Big|+\cdots + \Big| \frac{99}{98!}-\frac{100}{99!}\Big| \\ &= 1+1+2-\frac{100}{99!} \\ &=4-\frac{100}{99!}\end{align*}$
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  4. #4
    Newbie AgentSmith's Avatar
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    Re: Summation of a series

    Quote Originally Posted by sbhatnagar View Post
    $\displaystyle S_k$ is an infinite geometric series.

    1.Sum of an infinite geometric series is given by $\displaystyle S=\frac{a}{1-r}$

    therefore $\displaystyle S_k= \frac{\frac{k-1}{k!}}{1-\frac{1}{k}}$

    $\displaystyle \implies S_k = \frac{1}{(k-1)!}$

    2. Note That:
    $\displaystyle \begin{align*} \sum_{k=1}^{100}\Big| \frac{k^2-3k+1}{(k-1)!}\Big|&= \sum_{k=1}^{100}\Big| \frac{(k-1)^2-k}{(k-1)!}\Big| \\ &=\sum_{k=1}^{100}\Big| \frac{k-1}{(k-2)!}-\frac{k}{(k-1)!}\Big| \\ &= |-1|+|-1|+\Big| \frac{2}{1!}-\frac{3}{2!}\Big|+\Big| \frac{3}{2!}-\frac{4}{3!}\Big|+\cdots + \Big| \frac{99}{98!}-\frac{100}{99!}\Big| \\ &= 1+1+2-\frac{100}{99!} \\ &=4-\frac{100}{99!}\end{align*}$
    Thanks!

    That should mean my final answer is $\displaystyle 4-\frac{100}{99!}+\frac{100^2}{100!}$, but the answer provided to me is 4.
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  5. #5
    Member sbhatnagar's Avatar
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    Re: Summation of a series

    Quote Originally Posted by AgentSmith View Post
    Thanks!

    That should mean my final answer is $\displaystyle 4-\frac{100}{99!}+\frac{100^2}{100!}$, but the answer provided to me is 4.
    $\displaystyle \begin{align*}\frac{100^2}{100!}-\frac{100}{99!} &= \frac{100^2}{100!}-\frac{100}{99!}\cdot \frac{100}{100}\\ &=\frac{100^2}{100!}-\frac{100^2}{100!}\\&=0 \end{align*}$
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  6. #6
    Newbie AgentSmith's Avatar
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    Re: Summation of a series

    Quote Originally Posted by sbhatnagar View Post
    $\displaystyle \begin{align*}\frac{100^2}{100!}-\frac{100}{99!} &= \frac{100^2}{100!}-\frac{100}{99!}\cdot \frac{100}{100}\\ &=\frac{100^2}{100!}-\frac{100^2}{100!}\\&=0 \end{align*}$
    Thank you! I can't believe, I was being so stupid...
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