Question

Find ltn → ∞\(\sum_{a=0}^{n-1}\frac{sin(\frac{a}{n})}{n}\)

(a) ^1⁄a

(b) 1

(c) 1 – cos(1)

(d) 0

This question was posed to me in homework.

My doubt stems from Indeterminate Forms topic in chapter Differential Calculus of Engineering Mathematics

Answer 1

The correct answer is (c) 1 – cos(1)

To explain I would say: We use the concept of limit of a sum which is

  \(\int_a^b f(x)dx=lt_{n\rightarrow \infty}(\frac{b-a}{n})\times(f(a)+f(a+\frac{b-a}{n})….+f(a+\frac{(n-1)(b-a)}{n}))\)

Thus we have

\(\int_0^1 sin(x)dx=lt_{n\rightarrow \infty}(\frac{1}{n})\times(f(0)+f(\frac{1}{n})+….+f(\frac{n-1}{n}))\)

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

\(\int_0^1 sin(x)dx=lt_{n\rightarrow \infty}\sum_{a=0}^{n-1}\frac{sin(\frac{a}{n})}{n}\)

It is enough to evaluate the integral

\(\int_{0}^{1} sin(x)dx=[-cos(x)]_0^1\)=(cos(0)-cos(1))

=(1-cos(1))