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 sum compactly.
The expression is $\displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\dfrac{r^4}{r^5+n^5}$.
Step 2: Divide top and bottom by $n^5$.
Each term becomes $\dfrac{(r/n)^4}{1+(r/n)^5}\cdot\dfrac1n$.
Step 3: Recognise the Riemann sum.
This is $\displaystyle\lim_{n\to\infty}\dfrac1n\sum_{r=1}^{n}f\!\left(\dfrac{r}{n}\right)$ with $f(x)=\dfrac{x^4}{1+x^5}$, so it equals $\displaystyle\int_0^1\dfrac{x^4}{1+x^5}\,dx$.
Step 4: Substitute $u=1+x^5$.
Then $du=5x^4\,dx$, so $x^4\,dx=\dfrac{du}{5}$, and the limits go from $u=1$ to $u=2$.
Step 5: Integrate.
$\displaystyle\dfrac15\int_1^2\dfrac{du}{u}=\dfrac15\big[\log u\big]_1^2=\dfrac15\log 2$.
Step 6: Express as the option.
$\dfrac15\log 2=\log 2^{1/5}=\log\sqrt[5]{2}$.
\[ \boxed{\log\sqrt[5]{2}} \]