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I would greatly appreciate some help in finding $\lim \ a_n$ if $$a_n = \frac{1}{\sqrt[3]{n^3+1}} + \frac{1}{\sqrt[3]{n^3+2}}+\cdots+\frac{1}{\sqrt[3]{n^3+n}}.$$

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HINT:

Note that $$\dfrac1{\sqrt[3]{n^3+n}} \leq \dfrac1{\sqrt[3]{n^3+k}} \leq \dfrac1{\sqrt[3]{n^3+1}}$$ for all $k \in \{1,2,3,\ldots,n\}$.

Hence, $$\dfrac1{\sqrt[3]{1+1/n^2}} = \dfrac{n}{\sqrt[3]{n^3+n}} \leq \sum_{k=1}^{n} \dfrac1{\sqrt[3]{n^3+k}} \leq \dfrac{n}{\sqrt[3]{n^3+1}} = \dfrac1{\sqrt[3]{1+1/n^3}}$$

Now apply the squeeze/sandwich theorem.

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