Question

Find \(lt_{n\rightarrow\infty}\frac{(1^a+2^a+…+(n-1)^a)}{n^{a+1}}\)

(a) 1

(b) ^1⁄a + 1

(c) 0

(d) Undefined

This question was posed to me by my school principal while I was bunking the class.

My question is based upon Indeterminate Forms topic in division Differential Calculus of Engineering Mathematics

Answer 1

The correct option is (b) ^1⁄a + 1

Easy explanation: \(\int_0^1x^a dx=lt_{n\rightarrow\infty}\frac{1}{n}\times((\frac{1}{n})^a+(\frac{2}{n})^a+…+(\frac{n-1}{n})^a)\)

=\(lt_{n\rightarrow\infty}\frac{(1^a+2^a+…+(n-1)^a)}{n^{a+1}}\)

It is enough to evaluatethis integral

\(\int_0^1 x^adx=[\frac{x^{a+1}}{a+1}]_0^1\)

=\(\frac{1}{a+1}\)