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How to evaluate the following limit? $$\lim_{n \to \infty} \sum_{k=1}^n \frac{n+k}{n^3+k}$$

I think that every term of the sum is greater than the first one and smaller than the last one and then from the squeeze theorem the limit is $0$. However, I can't prove that inequality.

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    $\begingroup$ What are you asking? $\endgroup$ – Will M. Nov 19 '18 at 18:31
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    $\begingroup$ The sum is less than $n \frac{n +n}{n^3+1}$ $\endgroup$ – RRL Nov 19 '18 at 18:33
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This should work: $$\sum_{k=1}^n\frac{n+k}{n^3+k}\leq\sum_{k=1}^n\frac{n+k}{n^3}\leq\sum_{k=1}^n\frac{n+n}{n^3}=\frac{2n^2}{n^3}=\frac{2}{n}\to0$$

Also, every term is $\geq 0$ so $0$ is also a lower bound.

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  • $\begingroup$ Typo: $n + n = 2n$. $\endgroup$ – Paul Frost Dec 2 '18 at 10:25
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HINT:

Note that $n+1\le n+k\le 2n$ and $n^3+1\le n^3 +k\le n^3+n$.

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