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Math Help - Proving Limit

  1. #1
    Member SENTINEL4's Avatar
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    Proving Limit

    Prove that lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2}  {n^3+n}]=1

    I am thinking something like that :
    lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2}  {n^3+n}]= lim[\frac{n^2}{n^2(n+\frac{1}{n^2})}+\frac{n^2}{n^2(n+  \frac{2}{n^2})}+...+\frac{n^2}{n^2(n+\frac{1}{n})}] and since \frac{1}{n^2}\rightarrow0 we get
    lim[\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}]=1

    Am i right?
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  2. #2
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    Prove It's Avatar
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    What value are you making \displaystyle n approach?
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  3. #3
    Member SENTINEL4's Avatar
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    Quote Originally Posted by Prove It View Post
    What value are you making \displaystyle n approach?
    n\rightarrow\infty
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  4. #4
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    Quote Originally Posted by SENTINEL4 View Post
    Prove that lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2}  {n^3+n}]=1

    I am thinking something like that :
    lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2}  {n^3+n}]= lim[\frac{n^2}{n^2(n+\frac{1}{n^2})}+\frac{n^2}{n^2(n+  \frac{2}{n^2})}+...+\frac{n^2}{n^2(n+\frac{1}{n})}] and since \frac{1}{n^2}\rightarrow0 we get
    lim[\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}]=1

    Am i right?
    According to me your method is wrong...
    instead take (n+1/(n^2))...... in the numerator with an exponent of '-1' and take 'n' common from each bracket and then expand the bracket binomially....
    now apply the limit n->infinity..you should get the answer..
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  5. #5
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    Just because a term "goes to 0", that does't mean you can ignore it in the sum. Its values for finite n are NOT 0 and will affect the sum. For example, 1/n goes to 0 but \sum\frac{1}{n} goes to infinity.
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  6. #6
    Member SENTINEL4's Avatar
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    Which is the correct method to reach to the answer then?
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  7. #7
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    I don't think this limit does go to \displaystyle 1...

    The limit of a sum is the sum of the limits, and if you evaluate the limit of each term using L'Hospital's Rule, you should find each term \displaystyle \to 0.

    So the limit should be \displaystyle 0...
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  8. #8
    Member SENTINEL4's Avatar
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    Quote Originally Posted by SENTINEL4 View Post
    Prove that lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2}  {n^3+n}]=1
    That's exactly what the exercise in the book says.
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  9. #9
    MHF Contributor chisigma's Avatar
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    The problem is to find [if it exists...] the limit of the sequence...

    \displaystyle s_{n}= \sum_{k=1}^{n} a_{k}= \sum_{k=1}^{n} \frac{1}{n+\frac{k}{n^{2}}} = \sum_{k=1}^{n} \frac{1}{n} (1-\frac{k}{n^{3}+k}) (1)

    It is evident from (1) that, because s_{n} is the sum of n positive terms and for all k is a_{k}< \frac{1}{n} for all n will be s_{n}<1. But for all k is also...

    \displaystyle \lim_{n \rightarrow \infty} n\ a_{k}=1 (2)

    ... so that...

    \displaystyle \lim_{n \rightarrow \infty} s_{n}=1 (3)

    The first ten s_{n} are...

    s_{1}= .5

    s_{2}= .844444444444...

    s_{3}= .931773399015...

    s_{4}= .962678176374...

    s_{5}= .976681734034...

    s_{6}= .984114247235...

    s_{7}= .988505461555...

    s_{8}= .991307021026...

    s_{9}= .993200301061...

    s_{10}= .994538200011...

    Kind regards

    \chi \sigma
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    If L is the limit of (a_n) (the given sequence) we have:

    \dfrac{n^2}{n^3+n}+\ldots+\dfrac{n^2}{n^3+n}\leq a_n\leq \dfrac{n^2}{n^3}+\ldots+\dfrac{n^2}{n^3}

    Equivalently:

    \dfrac{n^3}{n^3+n}\leq a_n\leq \dfrac{n^3}{n^3}

    Taking limits:

    1\leq L\leq 1

    which implies L=1 .


    Fernando Revilla
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  11. #11
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Prove It View Post
    The limit of a sum is the sum of the limits ...

    Here the result is not valid because the number of terms of a_n depends on n .


    Fernando Revilla
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  12. #12
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    Quote Originally Posted by Prove It View Post
    I don't think this limit does go to \displaystyle 1...

    The limit of a sum is the sum of the limits, and if you evaluate the limit of each term using L'Hospital's Rule, you should find each term \displaystyle \to 0.

    So the limit should be \displaystyle 0...
    Here we cannot apply L'Hopital's rule since the function is not continuous as it takes only integer values .the function will have discreet set of points.
    For ex:if 1<=n<=2 it will take only n=1 and n=2 but not the real nos. b/w 1&2
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