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Math Help - Sigma Notation

  1. #1
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    Sigma Notation

    I'm sorry if you have a diffuclty reading this. I am not sure how i would go about displaying this in HTML

    well

    above the Sigma there is an (n) and below (i=1) on the right side of the sigma this is displayed (1+ 2i/n) (2/n). Let s(n) = the above problem

    Find the limit of s(n) as n -> infinity

    How do I solve this problem? Could someone help me out step by step please?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by radioheadfan View Post
    I'm sorry if you have a diffuclty reading this. I am not sure how i would go about displaying this in HTML

    well

    above the Sigma there is an (n) and below (i=1) on the right side of the sigma this is displayed (1+ 2i/n) (2/n). Let s(n) = the above problem

    Find the limit of s(n) as n -> infinity

    How do I solve this problem? Could someone help me out step by step please?
    \sum_{i=1}^n(1+\frac{2i}{n})(\frac{2}{n})=\frac{2}  {n}\left[\sum_{i=1}^n1+\frac{2}{n}\sum_{i=1}^ni\right]

    Can you take it from here?
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    \sum_{i=1}^n(1+\frac{2i}{n})(\frac{2}{n})=\frac{2}  {n}\left[\sum_{i=1}^n1+\frac{2}{n}\sum_{i=1}^ni\right]

    Can you take it from here?
    not to be a pain, but not quite
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by radioheadfan View Post
    not to be a pain, but not quite
    Well, do you see what I have done thus far?
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  5. #5
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    VonNemo19's expression is the result of factoring; the n is an upper limit, not an index, so it can come out. Now you should know both of these identities:

    \sum_{i=1}^n 1 = n (trivial), and

    \sum_{i=1}^n i = \frac{1}{2}n(n+1).

    The second identity can be proved by induction, or, better yet, by the ingenius method of Gauss, letting s be the sum:

    s = 1 + 2 + 3 + ... + n
    s = n + ... + 3 + 2 + 1

    Each line has n terms, and the terms of each line, added together, equal n+1. Hence 2s = n(n+1), or s= n(n+1)/2.

    Using these identities, you should be able to solve the problem easily, letting n approach infinity.
    Last edited by AlephZero; July 31st 2009 at 08:03 PM. Reason: typo
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  6. #6
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    I did the problem out and came out with the number 2, is this correct?
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by radioheadfan View Post
    I did the problem out and came out with the number 2, is this correct?
    4 is correct.

    \sum_{i=1}^n(1+\frac{2i}{n})(\frac{2}{n})

    You can see that \frac{2}{n} will be a factor of every term, so therefore we can pull it out in front of the sigma symbol

    \frac{2}{n}\sum_{i=1}^n(1+\frac{2i}{n})

    Now, understand that if I were to sum the remaining terms together, they would expand together. Heres what I mean...

    \sum_{i=1}^n(1+\frac{2i}{n})=(1+\frac{2(1)}{n})+(1  +\frac{2(2)}{n})+(1+\frac{2(3)}{n})+...+(1+\frac{2  (n)}{n})

    So, you can see how I could rearange the terms to be like this...

    \sum_{i=1}^n(1+\frac{2i}{n})=[1+1+1+...+1]+\left[\frac{2(1)}{n}+\frac{2(2)}{n}+\frac{2(3)}{n}+...+\  frac{2(n)}{n}\right]

    So, it's like I have two sums

    One is \sum_{i=1}^n1, and the other is \sum_{i=1}^n\frac{2i}{n}.

    So, then this gives \frac{2}{n}\sum_{i=1}^n(1+\frac{2i}{n})=\frac{2}{n  }\left[\sum_{i=1}^n1+\sum_{i=1}^n\frac{2i}{n}\right]

    Now notice the second term inside the brackets. I can factor \frac{2}{n} out

    \frac{2}{n}\sum_{i=1}^n(1+\frac{2i}{n})=\frac{2}{n  }\left[\sum_{i=1}^n1+\frac{2}{n}\sum_{i=1}^ni\right]

    Now from the formulas

    \sum_{i=1}^nk=nk. and \sum_{i=1}^ni=\frac{n(n+1)}{2} we get

    \frac{2}{n}\left[\sum_{i=1}^n1+\frac{2}{n}\sum_{i=1}^ni\right]=\frac{2}{n}\left[n+\frac{2}{n}\cdot\frac{n(n+1)}{2}\right]=\frac{2}{n}[n+(n+1)]=\frac{2}{n}(2n+1)=4+\frac{2}{n}

    Taking the limit

    \lim_{n\to\infty}\sum_{i=1}^n(1+\frac{2i}{n})(\fra  c{2}{n})=\lim_{n\to\infty}(4+\frac{2}{n})=4
    Last edited by VonNemo19; July 31st 2009 at 11:44 PM.
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  8. #8
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    omg I don't get it . Help please.
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  9. #9
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by radioheadfan View Post
    omg I don't get it . Help please.
    I updated my last post. I hope you find it helpful.
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  10. #10
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    Talking

    Quote Originally Posted by radioheadfan View Post
    omg I don't get it . Help please.
    Please specify where, in the complete worked solution, you are getting lost.

    Thank you!
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  11. #11
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    Here's a somewhat more compact working...

    \sum_{i=1}^n\left(1+\frac{2i}{n}\right)\left(\frac  {2}{n}\right)=\frac{2}{n}\left(\sum_{i=1}^n 1 + \frac{2}{n}\sum_{i=1}^n i\right) (factoring)

    =\frac{2}{n}\left[n+\left(\frac{2}{n}\cdot \frac{1}{2}n(n+1)\right)\right] (by the identities given in my previous post)

    =\frac{2}{n}\left(2n+1\right) (algebraic simplification)

    =4 + \frac{2}{n} \rightarrow 4 + 0 = 4 \text { as } n \rightarrow \infty (further simplification and taking the limit.)
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