Question

Evaluate: \[ \lim_{n\to\infty}\sum_{r=n}^{2n}\left(\frac{n^3+r^3}{n^4}\right) \]

• A. \(\displaystyle \frac{19}{4}\)
• B. \(\displaystyle \frac{21}{4}\)
• C. \(\displaystyle \frac{23}{4}\)
• D. \(\displaystyle \frac{27}{4}\)

Hint:

Expressions of the form \[ \sum \frac1n f\left(\frac{r}{n}\right) \] usually represent a definite integral.

Answer 1

The Correct Option is A

Concept: Large summations involving \(n\) are converted into definite integrals using Riemann sums.

Step 1: Simplify the summand. \[ \frac{n^3+r^3}{n^4} = \frac1n+\frac1n\left(\frac{r}{n}\right)^3 \] Thus, \[ S_n= \sum_{r=n}^{2n}\left[\frac1n+\frac1n\left(\frac{r}{n}\right)^3\right] \] \[ = \sum_{r=n}^{2n}\frac1n + \sum_{r=n}^{2n}\frac1n\left(\frac{r}{n}\right)^3 \]

Step 2: Evaluate the first summation. From \(r=n\) to \(2n\), total terms are \(n+1\). Hence, \[ \sum_{r=n}^{2n}\frac1n = \frac{n+1}{n} \to 1 \]

Step 3: Convert second sum into integral. \[ \sum_{r=n}^{2n}\frac1n\left(\frac{r}{n}\right)^3 \] is a Riemann sum for \[ \int_1^2 x^3\,dx \] Now, \[ \int_1^2 x^3dx = \left[\frac{x^4}{4}\right]_1^2 = \frac{16-1}{4} = \frac{15}{4} \] Therefore, \[ \lim S_n = 1+\frac{15}{4} = \frac{19}{4} \] Hence, \[ \boxed{\frac{19}{4}} \]