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    Hi all, can someone please guide me how to do this question? I have no idea how to approach it .

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    (Original post by akimbo)
    Hi all, can someone please guide me how to do this question? I have no idea how to approach it .

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    Draw a graph, and draw on the appropriate sum in terms of rectangles. What does that look like in comparison with the integral?
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    (Original post by Smaug123)
    Draw a graph, and draw on the appropriate sum in terms of rectangles. What does that look like in comparison with the integral?
    Ok thanks. Would this be correct? 1 + 1/4 + 1/9 +...+ 1/n^2 > -1/(n+1) + 1? I'm still kind of confused.
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    Where did you get your expression on the RHS from? (It's not right).
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    (Original post by DFranklin)
    Where did you get your expression on the RHS from? (It's not right).
    I integrated 1/x^2 between n+1 and 1 to get the area underneath the curve. Can you explain why it's wrong?
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    (Original post by akimbo)
    I integrated 1/x^2 between n+1 and 1 to get the area underneath the curve. Can you explain why it's wrong?
    Wrong sign (which you seem to have found).

    General comment: at this level, you should always be "sanity checking" your calculations. It should be obvious that \int_1^{n+1} \dfrac{1}{x^2}\,dx is going to be > 0, so your answer can't be negative.

    Other than that, it's fine.
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    (Original post by DFranklin)
    Wrong sign (which you seem to have found).

    General comment: at this level, you should always be "sanity checking" your calculations. It should be obvious that \int_1^{n+1} \dfrac{1}{x^2}\,dx is going to be > 0, so your answer can't be negative.

    Other than that, it's fine.
    Ok thanks, could you explain how I get from here to show that the statement is true?
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    Am I correct in saying that \sum_{k=1}^n 1/k^2 = 1 as n  \Rightarrow \infty?
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    What are the first few terms?
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    (Original post by akimbo)
    Am I correct in saying that \sum_{k=1}^n 1/k^2 = 1 as n  \Rightarrow \infty?
    No, since even if you take n=2, the sum is already bigger than 1 and it will only increase. This is one of Euler's big results: the value of the sum is in the spoiler.
    Spoiler:
    Show
    \frac{\pi^2}6

    It's a bit of a pain to prove, although it's easy to show that S, the value of the sum, satisfies S > 1 > S-1, by considering the integral of 1/x^2 from 1 to infinity.
    https://en.wikipedia.org/wiki/Basel_problem outlines a proper proof.
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    (Original post by Smaug123)
    No, since even if you take n=2, the sum is already bigger than 1 and it will only increase. This is one of Euler's big results: the value of the sum is in the spoiler.
    Spoiler:
    Show
    \frac{\pi^2}6

    It's a bit of a pain to prove, although it's easy to show that S, the value of the sum, satisfies S > 1 > S-1, by considering the integral of 1/x^2 from 1 to infinity.
    https://en.wikipedia.org/wiki/Basel_problem outlines a proper proof.
    Oh god I'm so confused. :confused:
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    (Original post by akimbo)
    Oh god I'm so confused. :confused:
    The question's only asking you to put an upper bound on the sum, not evaluate it

    You've virtually done it now: your sum is < 2 - (1/n) for any n > 1, so what will the upper bound be as n-> infinity?
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    (Original post by akimbo)
    Oh god I'm so confused. :confused:
    Ah, I said that "it's easy to show…" and then restated the question you've been given :P
    The value of the integral \int_1^{n}\frac{1}{x^2} dx is [latex]-\frac{1}{n}+1[\latex], yes? If you draw a graph of y=1/x^2 and draw in the usual bounding rectangles for the integral (above and below the curve) so that the bounds look like the sum you've been given, what can you then say about the sum?
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    (Original post by davros)
    The question's only asking you to put an upper bound on the sum, not evaluate it

    You've virtually done it now: your sum is < 2 - (1/n) for any n > 1, so what will the upper bound be as n-> infinity?
    Would it be 2? Or was that a stupid answer?
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    (Original post by akimbo)
    Would it be 2? Or was that a stupid answer?
    Quite right, it is 2
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    (Original post by Smaug123)
    Quite right, it is 2
    Ok thanks! Thanks everyone! Just to double check I am right I'll go through my working.

    Area underneath the curve between x=1 and x=n+1 is given by

    \int^{n+1}_1 \frac{1}{x^2}\dx = 1 - \frac{1}{n+1}

    [graphs showing upper and lower bounds]

    Figure 1 shows that the area of the dotted rectangles is greater than the area under the curve.

    Hence,

     1 + \frac{1}{4} + \frac{1}{9} +...+ \frac{1}{n^2} &gt; \frac{-1}{n+1} + 1

    Which is,

    \displaystyle\sum_{k=1}^n \frac{1}{k^2} &gt; 1 - \frac{1}{n+1}

    Figure 2 shows the area of the dashed rectangles is less than the area under the curve.

    Hence,

    \frac{1}{4} + \frac{1}{9} +...+ \frac{1}{n^2} &lt; 1 - \frac{1}{n}
    1 + \frac{1}{4} + \frac{1}{9} +...+ \frac{1}{n^2} &lt; 2 - \frac{1}{n}

    Therefore,
    \displaystyle\sum_{k=1}^n \frac{1}{k^2} &lt; 2 - \frac{1}{n+1}

    If \displaystyle\sum_{k=1}^n \frac{1}{k^2} &lt; 2 - \frac{1}{n} for any   n&gt;1 then the upper bound for the infinite series must 2 as n -> infinity.

    Would this be correct?
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    (Original post by akimbo)
    Would this be correct?
    It is correct, yes - it's probably my fault, but you've done more than is asked for now because you've also provided a lower bound of 1, in the "Figure 1" bit. (I know I said to draw in both sets of rectangles, but I suppose it never hurts to have a lower bound too!) You don't actually *need* to have derived \displaystyle\sum_{k=1}^n \frac{1}{k^2} &gt; 1 - \frac{1}{n+1} for this question.
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    (Original post by Smaug123)
    It is correct, yes - it's probably my fault, but you've done more than is asked for now because you've also provided a lower bound of 1, in the "Figure 1" bit. (I know I said to draw in both sets of rectangles, but I suppose it never hurts to have a lower bound too!) You don't actually *need* to have derived \displaystyle\sum_{k=1}^n \frac{1}{k^2} &gt; 1 - \frac{1}{n+1} for this question.
    Thanks a bunch! I'm just glad I've got an answer for this now, couldn't have been possible without the help from everyone in this thread! Many thanks!
 
 
 
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