Question

Evaluate \[ \lim_{n\to\infty} \frac{1+2^4+3^4+\cdots+n^4}{n^5} - \lim_{n\to\infty} \frac{1+2^3+3^3+\cdots+n^3}{n^5}. \]

• A. \[ \frac15 \]
• B. \[ \frac14 \]
• C. \[ \frac1{20} \]
• D. \[ 0 \]

Hint:

For limits involving \[ \frac{\sum k^p}{n^{p+1}}, \] use the standard result \[ \lim_{n\to\infty}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}} = \frac1{p+1}. \] Thus, \[ \frac{\sum k^4}{n^5}\to\frac15, \qquad \frac{\sum k^3}{n^5}\to0. \]

Answer 1

The Correct Option is A

Concept: Use the standard limits \[ \sum_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}, \] and \[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}2\right)^2. \]

Step 1: Evaluate the first limit. Let \[ L_1= \lim_{n\to\infty} \frac{\sum_{k=1}^{n}k^4}{n^5}. \] Using the formula, \[ L_1 = \lim_{n\to\infty} \frac{n(n+1)(2n+1)(3n^2+3n-1)} {30\,n^5}. \] Divide numerator and denominator by \(n^5\): \[ L_1 = \frac1{30} \lim_{n\to\infty} \left(1+\frac1n\right) \left(2+\frac1n\right) \left(3+\frac3n-\frac1{n^2}\right). \] \[ L_1 = \frac1{30}(1)(2)(3) = \frac15. \]

Step 2: Evaluate the second limit. Let \[ L_2= \lim_{n\to\infty} \frac{\sum_{k=1}^{n}k^3}{n^5}. \] Using \[ \sum_{k=1}^{n}k^3 = \left(\frac{n(n+1)}2\right)^2, \] \[ L_2 = \lim_{n\to\infty} \frac{n^2(n+1)^2}{4n^5}. \] \[ = \frac14 \lim_{n\to\infty} \frac{(n+1)^2}{n^3}. \] \[ = \frac14 \lim_{n\to\infty} \frac{\left(1+\frac1n\right)^2}{n}. \] \[ =0. \]

Step 3: Compute the required value. \[ L_1-L_2 = \frac15-0. \] \[ =\frac15. \]

Step 4: Write the final answer. \[ \boxed{\frac15} \]