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Math Help - Value of a series.

  1. #1
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    Smile Value of a series.

    find the sum
    1+1/√2 +1/√3+1√4------1/√n where n is perfect square
    also find greatest integer value of (1+1/√2 +1/√3+1√4------1/√n ) where n is perfect square
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  3. #3
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    Quote Originally Posted by ayushdadhwal View Post
    find the sum
    1+1/√2 +1/√3+1√4------1/√n where n is perfect square
    also find greatest integer value of (1+1/√2 +1/√3+1√4------1/√n ) where n is perfect square
    Let S_n = \sum_{r=1}^n\frac1{\sqrt r}. Then S_n = \sqrt n\Bigl(\frac1n\sum_{r=1}^n\frac1{\sqrt {r/n}}\Bigr), and the expression in parentheses is a lower Riemann sum for \displaystyle\int_0^1\!\!\!x^{-1/2}dx = 2. Therefore S_n<2\sqrt n. On the other hand, the expression in parentheses is an upper Riemann sum for \displaystyle\int_{1/n}^{(n+1)/n}\!\!\!x^{-1/2}dx, which leads to the estimate S_n>2\sqrt{n+1}-2.

    That's as close as I can get to pinning down the value of S_n. Its integer part (when n is a perfect square) must be either 2\sqrt n-2 or 2\sqrt n-1.
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    Second attempt at this problem. We are interested in the sum S_n = \sum_{r=1}^n\frac1{\sqrt r} when n is a perfect square, say n = k^2. Then the difference between successive sums of this form is

    \displaystyle S_{(k+1)^2} - S_{k^2} = \sum_{r=1}^{2k+1}\frac1{\sqrt{k^2+r}} = \sqrt{2k+1}\biggl(\frac1{2k+1}\sum_{r=1}^{2k+1}\fr  ac1{\sqrt {\alpha +\frac r{2k+1}}}\biggr),

    where \alpha = k^2/(2k+1). As in the previous comment, this is a lower Riemann sum for

    \displaystyle\sqrt{2k+1}\int_0^1\!\!\!(\alpha+x)^{-1/2}dx =  \sqrt{2k+1}\Bigl[2(\alpha+x)^{1/2}\Bigr]_0^1 = 2

    (because \sqrt{\alpha+1} - \sqrt\alpha = \frac{k+1}{\sqrt{2k+1}} - \frac k{\sqrt{2k+1}} = \frac1{\sqrt{2k+1}}). Therefore S_{(k+1)^2} - S_{k^2} < 2.

    The aim now is to show by induction that the integer part \lfloor S_{k^2}\rfloor of S_{k^2} is 2k-2. The base case is k=2, and for that you need to verify that S_4\approx2.78, so that \lfloor S_{4}\rfloor = 2.

    For the inductive step, suppose that \lfloor S_{k^2}\rfloor = 2k-2. The above argument shows that S_{(k+1)^2}<S_{k^2} + 2, so that \lfloor S_{(k+1)^2}\rfloor \leqslant \lfloor S_{k^2}\rfloor + 2 = 2(k+1)-2. The reverse inequality \lfloor S_{(k+1)^2}\rfloor \geqslant 2(k+1) - 2 comes from the previous comment, where I showed that S_n > 2\sqrt n - 2.

    Therefore the integer part of S_n, when n is a perfect square greater than 1, is given by \lfloor S_n\rfloor = 2\sqrt n - 2.
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    Thanks sir
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