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

Step 1: Split the term inside the sum.
\[ \frac{n^3+r^3}{n^4}=\frac1n+\frac1n\left(\frac{r}{n}\right)^3. \] So the sum breaks into two parts.
Step 2: Handle the first part.
From $r=n$ to $2n$ there are $n+1$ terms, each $\dfrac1n$, so the first part is $\dfrac{n+1}{n}\to1$ as $n\to\infty$.
Step 3: Recognise the second part as a Riemann sum.
$\displaystyle\sum_{r=n}^{2n}\frac1n\left(\frac{r}{n}\right)^3$ matches $\displaystyle\int_1^2 x^3\,dx$, because $\dfrac{r}{n}$ runs from $1$ to $2$ in steps of $\dfrac1n$.
Step 4: Evaluate the integral.
\[ \int_1^2 x^3\,dx=\left[\frac{x^4}{4}\right]_1^2=\frac{16-1}{4}=\frac{15}{4}. \]
Step 5: Add the two limits.
\[ \lim_{n\to\infty}S_n=1+\frac{15}{4}=\frac{19}{4}. \]
Step 6: State the answer.
The limit equals \[ \boxed{\frac{19}{4}}. \]