Question

If \(\lim_{n\to\infty}\dfrac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}\left[(na+1)+(na+2)+\cdots+(na+n)\right]}=\dfrac{1}{60}\) for some positive real number \(a\), then \(a\) is equal to :

• A. \(7\)
• B. \(8\)
• C. \(\dfrac{15}{2}\)
• D. \(\dfrac{17}{2}\)

Hint:

Numerator: \(\sum_{r=1}^n r^a = n^{a+1}\cdot\frac1n\sum \left(\frac rn\right)^a \to \frac{n^{a+1}}{a+1}\).
Bracket: \(\sum_{r=1}^{n}(na+r)=n\cdot na+\frac{n(n+1)}{2}=n^2\left(a+\frac12\right)+\frac n2\sim n^2\left(a+\frac12\right)\).
Denominator \(\sim n^{a-1}\cdot n^2\left(a+\frac12\right)=n^{a+1}\cdot\frac{2a+1}{2}\).
So the limit \(=\frac{1}{a+1}\cdot\frac{2}{2a+1}=\frac1{60}\), i.e. \((a+1)(2a+1)=120\).
\(2a^2+3a-119=0\Rightarrow a=\frac{-3\pm31}{4}\). Since \(a\gt0\), \(a=7\).
Check: \(\frac18\cdot\frac2{15}=\frac1{60}\).
Answer: (1) 7

Answer 1

The Correct Option is A

Numerator behaves like \(\frac{n^{a+1}}{a+1}\) (Riemann sum); the bracket is an arithmetic sum.
Equate the limit to \(\frac1{60}\) and solve the quadratic in \(a\).