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Math Help - series

  1. #1
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    Thumbs up series

    Dear Sir
    I would be grateful if you can help me to find f(r) in the below questions
    thanks
    Kingsman



    Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
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  2. #2
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    Quote Originally Posted by kingman View Post
    Dear Sir
    I would be grateful if you can help me to find f(r) in the below questions
    thanks
    Kingsman

    Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
    1)

    \displaystyle\huge\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}

    =\displaystyle\huge\frac{n+1}{(n+1)n!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}

    Therefore

    \displaystyle\huge\frac{1}{2!}+\frac{2}{3!}+......

    =1-\displaystyle\huge\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}+.......-\frac{1}{(n+1)!}

    =1-\displaystyle\huge\frac{1}{(n+1)!}
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  3. #3
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    Quote Originally Posted by kingman View Post
    Dear Sir
    I would be grateful if you can help me to find f(r) in the below questions
    thanks
    Kingsman


    Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
    2)

    \displaystyle\huge\frac{1}{(n+2)n!}=\frac{(n+1)}{(  n+2)(n+1)n!}=\frac{n+1}{(n+2)!}

    \displaystyle\huge\frac{n+2-1}{(n+2)!}=\frac{n+2}{(n+2)!}-\frac{1}{(n+2)!}

    \displaystyle\huge\frac{n+2}{(n+2)(n+1)!}-\frac{1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}

    Use this to form the telescope.
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  4. #4
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    Thanks very much for the kind help.
    May I know how and what is the idea which you employ in your solution making it so neat and simply. Is the f(r) unique and is there a general method to find f(r) ?.
    Lastly is it possible to find a g(r) such that the nth term= g(r)-g(r-1) allowing us to find the sum?
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  5. #5
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    If you are told to evaluate an infinite sum you should either change it into a known form of a function or you should make it a telescoping sum. Now to make it a telescoping sum you need to look for the following things:

    - an obvious way to apply partial fractions decomposition
    - add and subtract something in the numerator so that part of it looks like something that we can cancel on the bottom.

    That is usually a good start. There is a lot of intuition involved in some examples and less in others.
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  6. #6
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    Thanks very much for the reply.
    Can I know how to find g(r) such that the nth term= g(r)-g(r-1) thus allowing us to find the sum of the above 3 problem.
    How to apply telescope method in question 3?
    I cannot wrap around my head to get started in problem3
    Thanks
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  7. #7
    MHF Contributor chisigma's Avatar
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    3)

    The sum can be written as...

    \displaystyle \sum_{k=1}^{n} \frac{k}{k+5} =  n - 5\ \sum_{k=1}^{n} \frac{1}{k+5} (1)

    In order to solve the second term of (1) we have to introduce the function digamma, defined as...

    \displaystyle \psi(x)= \sum_{j=1}^{\infty} \frac{x}{j\ (j+x)} - \gamma (2)

    ... where \gamma is the 'Euler's constant'. From (2) You can derive the following basic identity...

    \displaystyle \psi(1+x) - \psi (x) = \frac{1}{1+x} (3)

    Now setting in (3) x=k+4 You obtain...

    \displaystyle \psi(k+5) - \psi (k+4) = \frac{1}{k+5} (4)

    ... so that is...

    \displaystyle \sum_{k=1}^{n} \frac{1}{k+5} =  \sum_{k=1}^{n} \{ \psi(k+5) - \psi (k+4) \} = \psi (n+5) - \psi(5) (5)

    ... and finally You can write...

    \displaystyle \sum_{k=1}^{n} \frac{k}{k+5} = n - 5\ \psi(n+5) + 5\ \psi(5) (6)

    For more information about the digamma function see here...

    Digamma function - Wikipedia, the free encyclopedia


    Kind regards

    \chi \sigma
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  8. #8
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    That's pretty cool, chisigma, but I think there is an even simpler answer.

    \displaystyle\sum_{k=1}^n \frac{k}{k+5} = \sum_{k=1}^n \frac{k+5-5}{k+5} = \sum_{k=1}^n \left(\frac{k+5}{k+5}-\frac{5}{k+5}\right)

    \displaystyle = \left(\sum_{k=1}^n 1\right)-5\left(\sum_{k=1}^n \frac{1}{k+5}\right) = n - 5\sum_{k=6}^{n+5} \frac{1}{k+5} = n - 5\left(H_n - \frac{1}{1}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right) = \frac{137}{12}+n-5H_n

    where [LaTeX ERROR: Convert failed] is the n-th Harmonic number. There is no simpler way to write this sum.
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