Results 1 to 6 of 6

Math Help - Summation of a series

  1. #1
    Newbie AgentSmith's Avatar
    Joined
    Dec 2011
    Posts
    12

    Summation of a series

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

    How do I do this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie AgentSmith's Avatar
    Joined
    Dec 2011
    Posts
    12

    Re: Summation of a series

    Quote Originally Posted by CaptainBlack View Post
    Have you considered starting by writing S_k as a function of k?
    CB
    Yes. I got S_k=\frac{1}{(k-1)!}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: Summation of a series

    S_k is an infinite geometric series.

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

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

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

    2. Note That:
    \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*}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie AgentSmith's Avatar
    Joined
    Dec 2011
    Posts
    12

    Re: Summation of a series

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

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

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

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

    2. Note That:
    \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 4-\frac{100}{99!}+\frac{100^2}{100!}, but the answer provided to me is 4.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: Summation of a series

    Quote Originally Posted by AgentSmith View Post
    Thanks!

    That should mean my final answer is 4-\frac{100}{99!}+\frac{100^2}{100!}, but the answer provided to me is 4.
    \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*}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie AgentSmith's Avatar
    Joined
    Dec 2011
    Posts
    12

    Re: Summation of a series

    Quote Originally Posted by sbhatnagar View Post
    \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...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Ad hoc summation of series
    Posted in the Number Theory Forum
    Replies: 26
    Last Post: January 14th 2011, 06:12 PM
  2. summation of a series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 17th 2009, 09:57 PM
  3. Summation of the series
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: July 23rd 2009, 04:56 AM
  4. Need help with summation/series.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 13th 2009, 07:51 PM
  5. Summation of a Series
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 14th 2008, 02:02 PM

Search Tags


/mathhelpforum @mathhelpforum