Problem 46 - Essential Calculus

Question

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{3}+1}} $$

Step-by-Step Solution

Answer

The infinite series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{3}+1}}\) converges.

Step 1: Denote the series to evaluate

Let's denote the series to evaluate as \(aₙ = \frac{1}{\sqrt{n^{3}+1}}\) for \(n >= 1\).

Step 2: Identify a comparison series

We want to compare our series to a simpler series that we know converges or diverges. Since the cube root grows faster than one, we can disregard it in the denominator and consider another series, \(bₙ = \frac{1}{\sqrt{n^{3}}}\) or \(bₙ = \frac{1}{n^{\frac{3}{2}}}\), for comparison.

Step 3: Apply the direct comparison test

Note that for all \(n >= 1\), we have \(0 ≤ aₙ ≤ bₙ\). Since the series \(\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}\) is a p-series with p = 3/2 > 1, it converges. Therefore, by the direct comparison test, our original series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{3}+1}}\) also converges.