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Chapter 27: Problem 625

Establish the convergence or divergence of the series: \([1 /(1+\sqrt{1})]+[1 /(1+\sqrt{2})]+[1 /(1+\sqrt{3})]+[1 /(1+\sqrt{4})]+\ldots\)

Short Answer

Expert verified
The given series has terms of the form \(\frac{1-\sqrt{n}}{1 - n}\), which can be majorized by the divergent harmonic series \(\frac{1}{n-1}\). Therefore, the given series \(\sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}}\) is also divergent.

Step by step solution

01

Simplify the terms

We are given the series: \[\sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}}\] First, let's simplify the terms by multiplying the numerator and denominator by the conjugate of the denominator: \[\frac{1}{1+\sqrt{n}} \cdot \frac{1-\sqrt{n}}{1-\sqrt{n}} = \frac{1-\sqrt{n}}{1 - n}\]
02

Find a comparison series

To apply the comparison test, we want to find a comparison series with terms that can be easily compared to the terms of our given series. Since the terms of our series have the form: \[\frac{1-\sqrt{n}}{1 - n}\] We can compare it to the series: \[\sum_{n=1}^{\infty} \frac{1}{n}\]
03

Apply the comparison test

We want to check whether \(\frac{1-\sqrt{n}}{1 - n}\) is less than, greater than, or equal to \(\frac{1}{n}\). For \(n \ge 2\), we have \(1-\sqrt{n} \le 0\), and \(1-n \le 0\). Therefore, the product of two negatives makes the terms of the given series positive. Also, since \(1-\sqrt{n} \ge -\sqrt{n}\) and \(1-n \le -n+1\), we can write: \[\frac{1-\sqrt{n}}{1-n} \le \frac{1}{n-1}\] This means that: \[\sum_{n=2}^{\infty} \frac{1-\sqrt{n}}{1 - n} \le \sum_{n=2}^{\infty} \frac{1}{n-1}\]
04

Evaluate the convergence of the comparison series

By inspection, we can see that the comparison series is a harmonic series with a shifted index: \[\sum_{n=2}^{\infty} \frac{1}{n-1} = \sum_{n=1}^{\infty} \frac{1}{n}\] The harmonic series is a well-known divergent series.
05

Conclude the convergence or divergence of the given series

Since the given series is majorized by a divergent series, the given series must also be divergent. Therefore, the series: \[\sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}}\] is divergent.

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Most popular questions from this chapter

Test the series: \(1+\left(2 ! / 2^{2}\right)+\left(3 ! / 3^{3}\right)+\left(4 ! / 4^{4}\right)+\ldots\) by means of the ratio test. If this test fails, use another test.

Establish the convergence or divergence of the series: \([1 /(1+\sqrt{1})]+[1 /(1+\sqrt{2})]+[1 /(1+\sqrt{3})]+[1 /(1+\sqrt{4})]+\ldots\)

Test the alternating series: \([(1+\sqrt{2}) / 2]-[(1+\sqrt{3}) / 4]+[(1+\sqrt{4}) / 6]-[(1+\sqrt{5}) / 8]+\ldots\) for convergence.

Find the limit of the sequence defined by \(\mathrm{x}_{1}=(2 / 3)\) and \(\mathrm{x}_{\mathrm{n}+1}=\left[\left(\mathrm{x}_{\mathrm{n}}+1\right) /\left(2 \mathrm{x}_{\mathrm{n}}+1\right)\right]\).

Determine if the series \((1 / 2)+(1 / 3)+\left(1 / 2^{2}\right)+\left(1 / 3^{2}\right)+\left(1 / 2^{3}\right)+\left(1 / 3^{3}\right)+\ldots .\) is convergent or divergent
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