<s>[INST] $\lim_{n\rightarrow \infty }(a_{0}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$ $a_{0},a_{1},...,a_{k}$ are real numbers and $a_{0}+a_{1}+...+a_{k}=0$
$$L=\lim_{n\rightarrow \infty }(a_{0}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$$
$L=?$
My attempt:
$a_{0}=-a_{1}-a_{2}-...-a_{k}$ so $L=\lim_{n\rightarrow \infty }[(-a_{1}-a_{2}-...-a_{k})\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k}]$
$L=\lim_{n\rightarrow \infty }(-a_{1}\sqrt[3]{n}-a_{2}\sqrt[3]{n}-...-a_{k}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$
Now I group the terms and I have
$L=\lim_{n\rightarrow \infty }[a_{1}(\sqrt[3]{n+1}-\sqrt[3]{n})+a_{2}(\sqrt[3]{n+2}-\sqrt[3]{n})+...+a_{k}(\sqrt[3]{n+k}-\sqrt[3]{n})]$
I split in more limits, the constants $a_{0},a_{1},...,a_{k}$ go ahead the limits and the limits are 0 after some calculations.In the end $L=a_{1}\cdot 0+a_{2}\cdot 0+...+a_{k}\cdot 0=0$
 [/INST] We may expand each term in the sum using the binomial theorem (or, equivalently, as a Taylor series):
$$\sqrt[3]{n+j}=\sqrt[3]{n}\,\left(1+\frac{j}{n}\right)^{1/3}\!\!=\sqrt[3]{n}\left(1+\frac{j}{3n}-\frac{j^2}{9n^2}+\cdots\right).$$
Therefore
$$S_n \doteq a_0 \sqrt[3]{n} + a_1 \sqrt[3]{n+1} + \cdots + a_k \sqrt[3]{n+k} = \\
\big(a_0 + a_1 + \cdots + a_k\big)\sqrt[3]{n} + \frac{0a_0 + 1a_1 + \cdots + ka_k}{3}n^{-2/3} + \cdots.$$
Because $a_0 + a_1 + \cdots + a_k = 0$, for fixed $k$ we have $S_n \propto n^{-2/3}$, so $L = 0$.
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