<s>[INST] Calculate the limit $\lim\limits_{n \to \infty}\left(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\right)$ I have to calculate the limit $$\lim_{n \to \infty}\left(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+4)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2}-2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\right)$$
I tried to use Sandwich Theorem like this $$\frac{3n^{2}}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}} \leq \Bigg(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+4)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2}-2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\Bigg) \leq \frac{3n^{2}}{\sqrt[3]{(8n^{3}+2)}^{2}}$$
And for result I got that limit is $\frac{3}{4}$
Is this correct?
 [/INST] You could have done it a bit faster and get more than the limit using generalized harmonic numbers.
$$S_n=\sum_{k=1}^{3n^2}\frac{1}{\left(8 n^3+2k\right)^{3/2}}=\frac 1{2\sqrt 2}\Bigg[H_{4 n^3+3 n^2}^{\left(\frac{3}{2}\right)}-H_{4
   n^3}^{\left(\frac{3}{2}\right)} \Bigg]$$
Now, using the asymptotics
$$H_{p}^{\left(\frac{3}{2}\right)}=\zeta \left(\frac{3}{2}\right)-\frac 2{p^{1/2}}+\frac  1{2p^{3/2}}-\frac  1{8p^{5/2}}+O\left(\frac{1}{p^{9/2}}\right)$$ apply it twice and continue with Taylor series to get
$$S_n=\frac{3}{16 \sqrt{2}\, n^{5/2}}\Bigg[1-\frac{9}{16 n}+\frac{45}{128 n^2}-\frac{1713}{4096
   n^3}+O\left(\frac{1}{n^4}\right) \Bigg]$$
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