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Math Help - 1/[ n*sqrt(n+1) + (n+1)*sqrt(n) ] finding sum of this serie

  1. #1
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    1/[ n*sqrt(n+1) + (n+1)*sqrt(n) ] finding sum of this serie

    Hello,

    I am trying to solve this series:

    \sum_{n=1}^{\infty}\frac{1}{n\sqrt(n+1)+(n+1)\sqrt  (n)}

    I have tried partial decomposition (telescoping series) process but it results very complicated and incorrect and then I get stack.

    Have anyone an idea how the partial decompostion should be done so I get a result that the terms start to canceling them self. I don't want to bother anymore, just that part is enough and I will continue with the rest of the solution.

    Thank you everyone!
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  2. #2
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    Re: 1/[ n*sqrt(n+1) + (n+1)*sqrt(n) ] finding sum of this serie

    You probably need to start by rationalising the denominator.

    \displaystyle \begin{align*} \frac{1}{n\sqrt{n + 1} + (n + 1)\sqrt{n}} &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{[n\sqrt{n + 1} + (n + 1)\sqrt{n}][n\sqrt{n + 1} - (n + 1)\sqrt{n}]} \\  &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{n^2(n + 1) - n(n + 1)^2}\\ &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{n(n + 1)[n - (n + 1)]} \\ &= \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{n(n + 1)} \\ &= \frac{(n + 1)\sqrt{n } - n\sqrt{n + 1}}{(\sqrt{n})^2(\sqrt{n + 1})^2}\end{align*}

    Now try the Partial Fractions decomposition.
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  3. #3
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    Re: 1/[ n*sqrt(n+1) + (n+1)*sqrt(n) ] finding sum of this serie

     \begin{aligned}& \sum_{1 \le n \le k}\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} = \sum_{1 \le n \le k}\bigg(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\bigg) = \sum_{1 \le n \le k}\frac{1}{\sqrt{n}}-\sum_{1 \le n \le k}\frac{1}{\sqrt{n+1}} \\& = \sum_{1 \le n+1 \le k}\frac{1}{\sqrt{n+1}}-\sum_{1 \le n \le k}\frac{1}{\sqrt{n+1}} = \sum_{0 \le n \le k-1}\frac{1}{\sqrt{n+1}}-\sum_{1 \le n \le k}\frac{1}{\sqrt{n+1}} \\& = 1-\frac{1}{\sqrt{k+1}}+\sum_{1 \le n \le k}\frac{1}{\sqrt{n+1}}-\sum_{1 \le n \le k}\frac{1}{\sqrt{n+1}} = 1-\frac{1}{\sqrt{k+1}}.\end{aligned}

    \begin{aligned}\therefore \sum_{n \ge 1}\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} = \lim_{k \to +\infty}\sum_{1 \le n \le k}\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} = \lim_{k \to +\infty}\bigg(1-\frac{1}{\sqrt{k+1}}\bigg) = 1.\end{aligned}
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    Re: 1/[ n*sqrt(n+1) + (n+1)*sqrt(n) ] finding sum of this serie

    Quote Originally Posted by Prove It View Post
    You probably need to start by rationalising the denominator.

    \displaystyle \begin{align*} \frac{1}{n\sqrt{n + 1} + (n + 1)\sqrt{n}} &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{[n\sqrt{n + 1} + (n + 1)\sqrt{n}][n\sqrt{n + 1} - (n + 1)\sqrt{n}]} \\  &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{n^2(n + 1) - n(n + 1)^2}\\ &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{n(n + 1)[n - (n + 1)]} \\ &= \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{n(n + 1)} \\ &= \frac{(n + 1)\sqrt{n } - n\sqrt{n + 1}}{(\sqrt{n})^2(\sqrt{n + 1})^2}\end{align*}
    Now try the Partial Fractions decomposition.
    What's with the last step? From the second last step, I think you can separate it into two fractions:

    \begin{aligned}\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)} = \frac{(n+1)\sqrt{n}}{n(n+1)}-\frac{n\sqrt{n+1}}{n(n+1)} = \frac{\sqrt{n}}{{n}}-\frac{\sqrt{n+1}}{n+1} = \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\end{aligned}.

    EDIT: now I realised that the last step might have been a hint towards breaking it into two fractions, as in above!
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  5. #5
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    Solved!

    Wow!

    Prove it: Among many other solutions I tried yours but resulted in a very complex denominator (got disappointed), yours is small and simple!
    coffe: well you completed it all.

    So, impressing! I am so greatful to you guys!

    Thank you very much!

    EDIT:
    That's the point coffeeMachine where I messed things up, I used the A + B method for partial decomposition when in contrary it is not needed because although it is not obvious the numerator helps as decompose the fraction right away. Very good that you mentioned it, excellent cooperation guys!
    Last edited by Melsi; June 18th 2011 at 11:38 AM.
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