Problem 42 - Essential Calculus

Question

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=2}^{\infty} \frac{1}{\sqrt{n}-1} $$

Step-by-Step Solution

Answer

The series \(\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}-1}\) diverges.

Step 1: Write down the given series

The given series is \(\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}-1}\).

Step 2: Identified a similar known series

The series \(\frac{1}{\sqrt{n}}\) or \(\sum_{n=2}^{\infty} \frac{1}{ \sqrt{n}}\) is identified. This series is known to diverge because it is a p-series with \(p = \frac{1}{2}\), which is less than 1.

Step 3: Direct Comparison Test

Now we need to check the condition of the Direct Comparison Test, \(\frac{1}{\sqrt{n}-1} \geq \frac{1}{\sqrt{n}}\), for all \(n \geq 2\), we find this to be true, since the denominator of \(\frac{1}{\sqrt{n}-1}\) is less than that of \(\frac{1}{\sqrt{n}}\), ensuring that the value of \(\frac{1}{\sqrt{n}-1}\) is larger.

Step 4: Conclusion

According to the Direct Comparison Test, if 0 ≤ aₙ ≤ bₙ for all n in some range, and if the series \(\sum_{n=2}^{\infty} bₙ\) diverges, then \(\sum_{n=2}^{\infty} aₙ\) also diverges. Here, \(aₙ = \frac{1}{\sqrt{n}-1}\) and \(bₙ = \frac{1}{\sqrt{n}}\). Since \(\sum_{n=2}^{\infty} bₙ\) diverges and \(aₙ \geq bₙ\) for all \(n \geq 2\), the given series \(\sum_{n=2}^{\infty} aₙ\) also diverges.