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Math Help - convergent sequence

  1. #1
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    convergent sequence

    I need to prove that this sequence is convergent:
    (an) = (1 + 1/(sqrt(2) + 1/sqrt(3) + .... + 1/sqrt(n) - 2sqrt(n)). I know that this sequence is decreasing and bounded below. Therefore, it converges. However, I'm not sure how to show it's lower bounded. It seems to be bounded below by -2. I can show its decreasing.

    Thanks for any help.
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  2. #2
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    The problem here is the sum,
    \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}}+... + \frac{1}{\sqrt{n}} .
    We have no "nice" formula for working with s_n.
    However, in analysis we should learn how to approximate.

    Here is how we do it.*

    Define the following function f(x) = \frac{1}{\sqrt{x}} (see below).
    Now the integral:
    \int_1^n \frac{dx}{\sqrt{x}}
    Is the area below this curve.
    In this particular case it is:
    \int_1^4 \frac{dx}{\sqrt{x}} = 2\sqrt{4+1} - 2
    And the area of the rectangles is (the width is 1):
    \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sq  rt{3}}
    Thus, we have:
    \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sq  rt{3}} \geq \int_1^4 \frac{dx}{\sqrt{x}} = 2\sqrt{4+1} - 2
    In general we have,
    \frac{1}{\sqrt{1}}+...+\frac{1}{\sqrt{n}} \geq \int_1^{n+1}\frac{dx}{\sqrt{x}} = 2\sqrt{n+1} - 2.
    Thus,
    \frac{1}{\sqrt{1}}+...+\frac{1}{\sqrt{n}} - 2\sqrt{n} \geq 2( \sqrt{n+1} -\sqrt{n}) - 2 \geq -2.



    *)The credit goes to the proof of the Integral Test for series.
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