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How to find $$ \lim_{n\rightarrow \infty} \frac {n^2}{n^3+n +1} + \frac {n^2}{n^3 +n +2} +.....+\frac {n^2}{n^3 + 2n}? $$

I was thinking about the Riemann sum, but I am not able to do it.

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    $\begingroup$ Sum of $n$ terms, each between $__$ and $__$ hence the limit is $__$. $\endgroup$
    – Did
    May 3, 2018 at 7:01
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    $\begingroup$ You should take a look here how to ask a good question math.meta.stackexchange.com/questions/9959/… $\endgroup$
    – user
    May 3, 2018 at 7:07
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    $\begingroup$ If the limit of $$\frac{n^3}{n^3+n+1}$$ is not obvious to you, as seems apparent from the comments below, this is the question you should be asking, rather than the one above. $\endgroup$
    – Did
    May 4, 2018 at 4:40

1 Answer 1

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\begin{align*} \sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+k}\leq\sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+1}=\dfrac{n^{3}}{n^{3}+n+1} \end{align*} and \begin{align*} \sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+k}\geq\sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+n}=\dfrac{n^{3}}{n^{3}+2n}, \end{align*} now use Squeeze Theorem to conclude.

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  • $\begingroup$ Thanks @user284331..can u more elaborate that how can i find $lim x_n$,,as i search on wolframalpha,,,it show that $\sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+k}$ is diverges ??? Pliz more elaborate ur answer.... $\endgroup$
    – user557742
    May 3, 2018 at 7:34
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    $\begingroup$ So by Squeeze Theorem the limit is $1$, I don't understand why Wolfram says that it is divergent. $\endgroup$
    – user284331
    May 3, 2018 at 7:35
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    $\begingroup$ W.A. is correct $n \to \infty $ sum is equal to $1$. Series is convergent. W.A. is correct. $\endgroup$
    – user548054
    May 3, 2018 at 7:53
  • $\begingroup$ @user284331 That mean according to ur answer $lim_{n\rightarrow \infty}\dfrac{n^{3}}{n^{3}+n+1} =1$,...Is it True/false ? pliz tell me $\endgroup$
    – user557742
    May 3, 2018 at 7:53
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    $\begingroup$ Yes, that is true. $\endgroup$
    – user284331
    May 3, 2018 at 7:56

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