Question

\[ \lim_{n\to\infty} \left( \frac{1}{1+n^5} + \frac{2^4}{2^5+n^5} + \frac{3^4}{3^5+n^5} +\cdots+ \frac{n^4}{n^5+n^5} \right) = \]

• A. \(\dfrac{1}{5}\log 3\)
• B. \(\dfrac{1}{3}\log 5\)
• C. \(\dfrac{1}{2}\log 5\)
• D. \(\log \sqrt[5]{2}\)

Hint:

To evaluate limits involving sums, try converting them into a Riemann integral using \[ \sum f\left(\frac{r}{n}\right)\frac{1}{n} \to \int f(x)\,dx \] as \(n\to\infty\).

Answer 1

The Correct Option is D

Step 1: Write the given summation in sigma notation.
The given expression is \[ \lim_{n\to\infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5} \] We simplify each term by dividing numerator and denominator by \(n^5\).
\[ = \lim_{n\to\infty} \sum_{r=1}^{n} \frac{\dfrac{r^4}{n^5}} {\dfrac{r^5}{n^5}+1} \] \[ = \lim_{n\to\infty} \sum_{r=1}^{n} \frac{1}{n} \cdot \frac{\left(\dfrac{r}{n}\right)^4} {1+\left(\dfrac{r}{n}\right)^5} \]

Step 2: Convert into a definite integral.
This is a Riemann sum of the form \[ \lim_{n\to\infty} \sum_{r=1}^{n} f\left(\frac{r}{n}\right)\frac{1}{n} = \int_0^1 f(x)\,dx \] where \[ f(x)=\frac{x^4}{1+x^5} \] Therefore, \[ \lim_{n\to\infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5} = \int_0^1 \frac{x^4}{1+x^5}\,dx \]

Step 3: Evaluate the integral.
Let \[ u=1+x^5 \] Then, \[ du=5x^4\,dx \] or \[ x^4\,dx=\frac{du}{5} \] Changing the limits: When \(x=0\), \[ u=1 \] When \(x=1\), \[ u=2 \] Thus, \[ \int_0^1 \frac{x^4}{1+x^5}\,dx = \frac{1}{5}\int_1^2 \frac{du}{u} \] \[ = \frac{1}{5}\left[\log u\right]_1^2 \] \[ = \frac{1}{5}(\log2-\log1) \] Since \(\log1=0\), \[ = \frac{1}{5}\log2 \] Using logarithmic property, \[ \frac{1}{5}\log2 = \log2^{1/5} \] \[ = \log\sqrt[5]{2} \]

Step 4: Final Answer.
Therefore, \[ \boxed{\log\sqrt[5]{2}} \]