/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Use the Direct Comparison Test t... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

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

Step by step solution

01

Write down the given series

The given series is \(\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}-1}\).
02

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.
03

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.
04

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.

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