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Thread: Proving Limit

  1. #1
    Member SENTINEL4's Avatar
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    Proving Limit

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

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

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

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

    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 $\displaystyle \sum\frac{1}{n}$ 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 $\displaystyle \displaystyle 1$...

    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 $\displaystyle \displaystyle \to 0$.

    So the limit should be $\displaystyle \displaystyle 0$...
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  8. #8
    Member SENTINEL4's Avatar
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    Quote Originally Posted by SENTINEL4 View Post
    Prove that $\displaystyle lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2} {n^3+n}]=1$
    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...

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

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

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

    ... so that...

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

    The first ten $\displaystyle s_{n}$ are...

    $\displaystyle s_{1}= .5$

    $\displaystyle s_{2}= .844444444444...$

    $\displaystyle s_{3}= .931773399015...$

    $\displaystyle s_{4}= .962678176374...$

    $\displaystyle s_{5}= .976681734034...$

    $\displaystyle s_{6}= .984114247235...$

    $\displaystyle s_{7}= .988505461555...$

    $\displaystyle s_{8}= .991307021026...$

    $\displaystyle s_{9}= .993200301061...$

    $\displaystyle s_{10}= .994538200011...$

    Kind regards

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

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

    Equivalently:

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

    Taking limits:

    $\displaystyle 1\leq L\leq 1$

    which implies $\displaystyle L=1$ .


    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 $\displaystyle a_n$ depends on $\displaystyle n$ .


    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 $\displaystyle \displaystyle 1$...

    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 $\displaystyle \displaystyle \to 0$.

    So the limit should be $\displaystyle \displaystyle 0$...
    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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