Question

Find the value of \[ \lim_{n\to\infty}\left(\frac{1^2}{n^3+1^3}+\frac{2^2}{n^3+2^3}+\cdots+\frac{1}{2n}\right) \] =

• A. \(\log 2\)
• B. \(2\log 2\)
• C. \(\dfrac{1}{2}\log 2\)
• D. \(\log \sqrt[3]{2}\)

Hint:

For limits of sums, try to express the term in the form \(\frac{1}{n}f\left(\frac{r}{n}\right)\), then convert it into a definite integral over \([0,1]\).

Answer 1

The Correct Option is D

Step 1: Write the general term of the sum.
The given expression can be written as \[ \lim_{n\to\infty}\sum_{r=1}^{n}\frac{r^2}{n^3+r^3} \] For \(r=n\), \[ \frac{n^2}{n^3+n^3}=\frac{n^2}{2n^3}=\frac{1}{2n} \] which matches the last term.

Step 2: Convert the expression into Riemann sum form.
Now, \[ \frac{r^2}{n^3+r^3} = \frac{r^2}{n^3\left(1+\frac{r^3}{n^3}\right)} \] \[ = \frac{1}{n}\cdot \frac{\left(\frac{r}{n}\right)^2}{1+\left(\frac{r}{n}\right)^3} \] Therefore, \[ \lim_{n\to\infty}\sum_{r=1}^{n}\frac{r^2}{n^3+r^3} = \lim_{n\to\infty}\sum_{r=1}^{n} \frac{1}{n}\cdot \frac{\left(\frac{r}{n}\right)^2}{1+\left(\frac{r}{n}\right)^3} \]

Step 3: Change the Riemann sum into definite integral.
Using the standard result, \[ \lim_{n\to\infty}\sum_{r=1}^{n}\frac{1}{n}f\left(\frac{r}{n}\right) = \int_0^1 f(x)\,dx \] Here, \[ f(x)=\frac{x^2}{1+x^3} \] Hence, \[ \lim_{n\to\infty}\sum_{r=1}^{n}\frac{r^2}{n^3+r^3} = \int_0^1 \frac{x^2}{1+x^3}\,dx \]

Step 4: Evaluate the integral.
Let \[ u=1+x^3 \] Then, \[ du=3x^2\,dx \] So, \[ x^2\,dx=\frac{du}{3} \] Now, \[ \int_0^1 \frac{x^2}{1+x^3}\,dx = \frac{1}{3}\int_{1}^{2}\frac{1}{u}\,du \] \[ = \frac{1}{3}\left[\log u\right]_{1}^{2} \] \[ = \frac{1}{3}\log 2 \]

Step 5: Express the answer in option form.
Since, \[ \frac{1}{3}\log 2=\log 2^{1/3} \] Therefore, \[ \frac{1}{3}\log 2=\log \sqrt[3]{2} \]

Step 6: Final conclusion.
Hence, \[ \boxed{\log \sqrt[3]{2}} \]