Limits, Continuity & Differentiability
Limit of Sum Formulas
Grade 12
Question:
<p>Find $$\lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + \ldots + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + 3^3 + \ldots + n^3}{n^5}$$</p>
<p>(a) $$\frac{1}{30}$$</p>
<p>(b) $$0$$</p>
<p>(c) $$\frac{1}{4}$$</p>
<p>(d) $$\frac{1}{5}$$</p>
Step-by-Step Solution
Key Concept: Use the standard formulas for sum of fourth and third powers, then divide by n^5 and take limits. The dominant term in each sum determines the limit's behavior.
Step 1: Recall the formulas for power sums
The standard formulas are:
$$\sum_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
$$\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}$$
Step 2: Evaluate the first limit
$$\lim_{n \to \infty} \frac{\sum_{k=1}^{n} k^4}{n^5} = \lim_{n \to \infty} \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}$$
Expanding the numerator, the leading term is $n \cdot n \cdot 2n \cdot 3n^2 = 6n^5$.
$$= \lim_{n \to \infty} \frac{6n^5 + \text{lower order terms}}{30n^5} = \frac{6}{30} = \frac{1}{5}$$
Step 3: Evaluate the second limit
$$\lim_{n \to \infty} \frac{\sum_{k=1}^{n} k^3}{n^5} = \lim_{n \to \infty} \frac{n^2(n+1)^2}{4n^5}$$
$$= \lim_{n \to \infty} \frac{n^2(n^2+2n+1)}{4n^5} = \lim_{n \to \infty} \frac{n^4 + \text{lower order terms}}{4n^5}$$
Since the degree of the numerator (4) is less than the degree of the denominator (5), the limit is 0.
$$= 0$$
Step 4: Calculate the difference
Subtract the second limit from the first limit:
$$\lim_{n \to \infty} \frac{\sum k^4}{n^5} - \lim_{n \to \infty} \frac{\sum k^3}{n^5} = \frac{1}{5} - 0 = \frac{1}{5}$$
Correct Answer: a