Question

For a $\in $ ℝ, $\left|a\right| > 1,$ let
$ \displaystyle\lim_{n \to \infty} \left(\frac{1+\sqrt[3]{2}+\cdots+\sqrt[3]{n}}{n^{7/3}\left(\frac{1}{\left(an+1\right)^{2}}+\frac{1}{\left(an+2\right)^{2}}+\cdots+\frac{1}{\left(an+n\right)^{2}}\right)}\right) = 54.$
Then the possible value(s) of a is/are

• A. -9
• B. -6
• C. 7
• D. 8

Answer 1

Answer: (A), (D)

$\displaystyle \lim_{n \to \infty} $$\frac{ n^{1/3}\left(\displaystyle \sum_{r=1}^n\left(\frac{r}{n}\right)^{1/3}\right)}{n^{7/3}\left(\displaystyle \sum_{r=1}^n\frac{1}{\left(an+r\right)^{2}}\right)}=54\Rightarrow $$\displaystyle \lim_{n \to \infty} $ $\left(\frac{\frac{1}{n}\displaystyle \sum_{r=1}^n\left(\frac{r}{n}\right)^{1/3}}{\frac{1}{n}\displaystyle \sum_{r=1}^n\frac{1}{\left(a+r/n\right)^{2}}}\right)=54$ $\Rightarrow $ $\frac{\int\limits^{{1}}_{{0}}x^{1/3}dx }{\int\limits^{{1}}_{{0}}\frac{1}{\left(a+x\right)^{2}}dx }=54$ $\Rightarrow \frac{\frac{3}{4}}{\frac{1}{a\left(a+1\right)}}=54$
$\Rightarrow a\left(a+1\right)=72 \Rightarrow a^{2}+a-72=0 \Rightarrow a=-9, 8$