Question

Let \[\mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{1 + \sqrt[3]{2} + ... + \sqrt[3]{n}}}{{{n^{\dfrac{7}{3}}}\left( {\dfrac{1}{{{{\left( {an + 1} \right)}^2}}} + \dfrac{1}{{{{\left( {an + 2} \right)}^2}}} + ... + \dfrac{1}{{{{\left( {an + n} \right)}^2}}}} \right)}}} \right) = 54\], for \[a \in R\], \[\left| a \right| > 1\]. Then what are the possible values of \[a\] ?
A. 8
B. \[ - 9\]
C. \[ - 6\]
D. 7

Answer 1

Hint: First, simplify the numerator and denominator on the lefthand side of the given equation of a limit as the summation. Then convert the summations of the numerator and denominator into separate integrals. After that, solve both integrals and get the equation with variable \[a\]. In the end, solve the equation to reach the required answer.

Formula used:
\[\int\limits_a^b {{x^n}dx} = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]_a^b\]
\[\int\limits_a^b {\dfrac{1}{{{x^2}}}dx} = \left[ {\dfrac{{ - 1}}{x}} \right]_a^b\]
\[{a^{m + n}} = {a^m}{a^n}\]

Complete step by step solution:
The given limit is \[\mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{1 + \sqrt[3]{2} + ... + \sqrt[3]{n}}}{{{n^{\dfrac{7}{3}}}\left( {\dfrac{1}{{{{\left( {an + 1} \right)}^2}}} + \dfrac{1}{{{{\left( {an + 2} \right)}^2}}} + ... + \dfrac{1}{{{{\left( {an + n} \right)}^2}}}} \right)}}} \right) = 54\], where \[a \in R\], and \[\left| a \right| > 1\].
Let’s solve the left-hand side of the above equation of the limit.
\[\mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{\sqrt[3]{1} + \sqrt[3]{2} + ... + \sqrt[3]{n}}}{{{n^{\dfrac{7}{3}}}\left( {\dfrac{1}{{{{\left( {an + 1} \right)}^2}}} + \dfrac{1}{{{{\left( {an + 2} \right)}^2}}} + ... + \dfrac{1}{{{{\left( {an + n} \right)}^2}}}} \right)}}} \right) = 54\]
Rewrite the numerator as the summation of \[n\] terms.
\[\mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{\sum\limits_{r = 1}^n {{{\left( r \right)}^{\dfrac{1}{3}}}} }}{{{n^{\dfrac{7}{3}}}\left( {\dfrac{1}{{{{\left( {an + 1} \right)}^2}}} + \dfrac{1}{{{{\left( {an + 2} \right)}^2}}} + ... + \dfrac{1}{{{{\left( {an + n} \right)}^2}}}} \right)}}} \right) = 54\]
Simplify \[{n^{\dfrac{7}{3}}}\] using the exponent property \[{a^{m + n}} = {a^m}{a^n}\]
\[\mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{\sum\limits_{r = 1}^n {{{\left( {\dfrac{r}{n}} \right)}^{\dfrac{1}{3}}}\dfrac{1}{n}} }}{{n\left( {\dfrac{1}{{{{\left( {an + 1} \right)}^2}}} + \dfrac{1}{{{{\left( {an + 2} \right)}^2}}} + ... + \dfrac{1}{{{{\left( {an + n} \right)}^2}}}} \right)}}} \right) = 54\]
\[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{\sum\limits_{r = 1}^n {{{\left( {\dfrac{r}{n}} \right)}^{\dfrac{1}{3}}}\dfrac{1}{n}} }}{{\dfrac{1}{n}\left( {\dfrac{{{n^2}}}{{{{\left( {an + 1} \right)}^2}}} + \dfrac{{{n^2}}}{{{{\left( {an + 2} \right)}^2}}} + ... + \dfrac{{{n^2}}}{{{{\left( {an + n} \right)}^2}}}} \right)}}} \right) = 54\]
Rewrite the denominator as the summation of \[n\] terms.
\[\mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{\sum\limits_{r = 1}^n {{{\left( {\dfrac{r}{n}} \right)}^{\dfrac{1}{3}}}\dfrac{1}{n}} }}{{\sum\limits_{r = 1}^n {\dfrac{{{n^2}}}{{{{\left( {an + r} \right)}^2}}}\dfrac{1}{n}} }}} \right) = 54\]
Simplify the above equation.
\[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{\sum\limits_{r = 1}^n {{{\left( {\dfrac{r}{n}} \right)}^{\dfrac{1}{3}}}\dfrac{1}{n}} }}{{\sum\limits_{r = 1}^n {\dfrac{1}{{{{\left( {a + \dfrac{r}{n}} \right)}^2}}}\dfrac{1}{n}} }}} \right) = 54\]
Now convert the above equation of a limit into an integral.
Let consider \[\dfrac{r}{n} = x\] and then in integration form \[\dfrac{1}{n}dr = dx\] and limits will be 0 and 1.
\[\dfrac{{\int\limits_0^1 {{{\left( x \right)}^{\dfrac{1}{3}}}dx} }}{{\int\limits_0^1 {\dfrac{1}{{{{(a + x)}^2}}}dx} }} = 54\]
Now integrate the integrals with respect to the variable \[x\] .
Apply the integration properties \[\int\limits_a^b {{x^n}dx} = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]_a^b\] and \[\int\limits_a^b {\dfrac{1}{{{x^2}}}dx} = \left[ {\dfrac{{ - 1}}{x}} \right]_a^b\].
\[\dfrac{{\left[ {\dfrac{{{x^{\dfrac{1}{3} + 1}}}}{{\dfrac{1}{3} + 1}}} \right]_0^1}}{{\left[ {\dfrac{{ - 1}}{{a + x}}} \right]_0^1}} = 54\]
Simplify the above equation.
\[\dfrac{{\left[ {\dfrac{{{x^{\dfrac{4}{3}}}}}{{\dfrac{4}{3}}}} \right]_0^1}}{{\left[ {\dfrac{{ - 1}}{{a + x}}} \right]_0^1}} = 54\]
\[ \Rightarrow \dfrac{{\dfrac{3}{4}\left[ {{x^{\dfrac{4}{3}}}} \right]_0^1}}{{\left[ {\dfrac{{ - 1}}{{a + x}}} \right]_0^1}} = 54\]
Now apply the upper and lower limits.
\[\dfrac{{\dfrac{3}{4}\left[ {{1^{\dfrac{4}{3}}} - 0} \right]}}{{\left[ {\dfrac{{ - 1}}{{a + 1}} - \dfrac{{ - 1}}{{a + 0}}} \right]}} = 54\]
\[ \Rightarrow \dfrac{{\dfrac{3}{4}\left[ 1 \right]}}{{\left[ {\dfrac{1}{a} - \dfrac{1}{{a + 1}}} \right]}} = 54\]
\[ \Rightarrow \dfrac{{\dfrac{3}{4}}}{{\left[ {\dfrac{1}{a} - \dfrac{1}{{a + 1}}} \right]}} = 54\]
Cross multiplies the above terms.
\[\dfrac{3}{{4 \times 54}} = \dfrac{1}{a} - \dfrac{1}{{a + 1}}\]
\[ \Rightarrow \dfrac{1}{{72}} = \dfrac{{a + 1 - a}}{{a\left( {a + 1} \right)}}\]
\[ \Rightarrow \dfrac{1}{{72}} = \dfrac{1}{{a\left( {a + 1} \right)}}\]
Cross multiplies the above terms.
\[{a^2} + a = 72\]
\[ \Rightarrow {a^2} + a - 72 = 0\]
Factorize the above equation.
\[{a^2} + 9a - 8a - 72 = 0\]
\[ \Rightarrow a\left( {a + 9} \right) - 8\left( {a + 9} \right) = 0\]
\[ \Rightarrow \left( {a + 9} \right)\left( {a - 8} \right) = 0\]
\[ \Rightarrow \left( {a + 9} \right) = 0\] or \[\left( {a - 8} \right) = 0\]
\[ \Rightarrow a = - 9\] or \[a = 8\]
Hence the correct options are A and B.

Note: Students often get confused about the integration of \[\int {\dfrac{1}{{{{(a + x)}^2}}}dx} \] .
Solve this integration by using the u-substitution method. Substitute \[a + x = u\] in the above integral and convert it as integral in terms of \[u\]. Then apply the rule \[\int {{u^n}du} = \dfrac{{{u^{n + 1}}}}{{n + 1}}\] with \[n = - 2\]. Solve the integral and resubstitute the value of \[u\] to get the value of the integral.