Problem 59 - Essential Calculus

Question

Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{n^{k-1}}{n^{k}+1}, \quad k>2 $$

Step-by-Step Solution

Answer

The series \( \frac{n^{k-1}}{n^{k}+1} \) converges.

Step 1: Choose a Comparable Series

Choose a series that is easy to compare with the given series. In this case, choose \( b_n = \frac{1}{n^2} \).

Step 2: Calculate the Limit

Calculate the limit as \( n \) approaches infinity of \( \frac{a_n}{b_n} \). We have,\[ \lim_{n\to\infty} \frac{n^{k-1}}{n^{k}+1} \cdot n^2 = \lim_{n\to\infty} \frac{n^{k-1+2}}{n^{k}+1} = \lim_{n\to\infty} \frac{n^{k+1}}{n^{k}+1}= 0 \] where \( a_n = \frac{n^{k-1}}{n^{k}+1} \) and \( b_n = \frac{1}{n^2} \).

Step 3: Apply the Limit Comparison Test

Applying the limit comparison test, since the limit is 0 and \( k > 2 \), scenarios wherein the limit as \( n \) approaches infinity of \( \frac{a_n}{b_n} \) is some finite number (>0) or zero implies that both \( a_n \) and \( b_n \) will either both converge or both diverge. Since \( b_n = \frac{1}{n^2} \) is a p-series with \( p = 2 > 1 \), then \( b_n \) is convergent. Therefore, \( a_n = \frac{n^{k-1}}{n^{k}+1} \) is also convergent.