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 formulas for sums of powers: \(\sum k^4\) and \(\sum k^3\), divide by \(n^5\), and take limits as \(n\to\infty\).

Step 1: \(L_1 = \lim_{n\to\infty} \frac{\sum k^4}{n^5}\). \(\sum k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\). Divide by \(n^5\): leading term \(\frac{1\cdot1\cdot2\cdot3}{30} = \frac{6}{30} = \frac15\). So \(L_1 = 1/5\).

Step 2: \(L_2 = \lim_{n\to\infty} \frac{\sum k^3}{n^5}\). \(\sum k^3 = [n(n+1)/2]^2\). Divided by \(n^5\): \(\sim \frac{n^4/4}{n^5} = \frac{1}{4n} \to 0\). So \(L_2 = 0\).

Step 3: \(L_1 - L_2 = 1/5\).

Step 4: Write the final answer. \(\boxed{\frac15}\)