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Thread: Ramanujan's Impossible sum?

  1. #1
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    Ramanujan's Impossible sum?

    Hi, I am in desperate need of some help with my final maths assignment for one of my maths subjects. I have tried for ages trying to work out this problem but just have no idea what to do!?! I have asked numerous people, but they dont know what to do. So I'm begging for some help!!!
    Here is the question:-

    Watch the Mathologer video Ramanujan: Making sense of 1+2+3+... = -1/12 and Co. https://www.youtube.com/watch?v=jcKR...ature=youtu.be
    a) In part 1 of the video it is shown how two infinite series can be multiplied. This wayof multiplying is called the Cauchy product. It turns out that the Cauchy product of twoabsolutely convergent series is convergent and its sum is the product of the sums of the twoseries that get multiplied. Use this information to answer the following question.What infinite series do you arrive at when you square the geometric series1 +1/2+1/4+1/8+ ∑ ∑ ∑?
    b) What is the sum of this new series?
    c) In part 3 of the video the Cesaro sum of a series is introduced. Calculate the Cesaro sumsof the series:
    1 − 1 + 0 + 1 − 1 + 0 + 1 − 1 + 0 + 1 − 1 + 0 + ∑ ∑ ∑
    1 − 1 + 0 + 0 + 1 − 1 + 0 + 0 + 1 − 1 + 0 + 0 + ∑ ∑ ∑ Both series are generated from the series 1 −1 + 1−1 +∑ ∑ ∑ by inserting 0s into this sum. Asyou will see, by doing so the Cesaro sum changes. On the other hand, given a convergent (inthe standard way) series, will itís sum change if you insert 0s into it? Explain.
    d) Guess the answer to the first of the hard homework assignments in part 4 of the video,that is, sum the series1 − 22 + 32 − 42 + 52 − ∑ ∑ ∑based on the sequence of averages of averages of averages of the partial sums of this series.Base your guess on the first 5000 elements of this sequence calculated using Mathematica.Too devilish? Okay, here is some Mathematica code (just copy and paste into a Mathematicanotebook) that you can modify and extend to calculate those 5000 elements (you only have torecord the last few of those 5000 elements in your write-up). It calculates and prints the first5000 elements of the sequence of averages of the partial sums of the series 1−1+ 1−1+∑ ∑ ∑ .

    I have watched the video multiple times and been attempting the questions for a good couple of days, and i am in desperate need of help!!! Thank you
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  2. #2
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    Re: Ramanujan's Impossible sum?

    \begin{array}{ c | c c c c c } & 1 & \tfrac12 & \tfrac14 & \tfrac18 & \cdots \\ \hline 1 & 1 & \tfrac12 & \tfrac14 & \tfrac18 & \cdots \\ \tfrac12 & \tfrac12 & \tfrac14 & \tfrac18 & \tfrac1{16} & \cdots \\ \tfrac14 & \tfrac14 & \tfrac18 & \tfrac1{16} & \tfrac1{32} & \cdots \\ \tfrac18 & \tfrac18 & \tfrac1{16} & \tfrac1{32} & \tfrac1{64} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array}

    So the diagonal sum is 1 + \tfrac22 + \tfrac34 + \tfrac48 + \tfrac5{16} + \ldots

    You should be able to do something with that (you should know the formula for the geometric series - what do you get if you differentiate the series?).

    What do you have for the remainder of the questions?
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