Question

Show that for testing the convergence of of the series $$ x+y+x^{2}+y^{2}+x^{3}+y^{3}+\cdots $$ where \(0

Short answer

The D'Alembert ratio test is inconclusive due to varying outputs for even and odd terms. The Cauchy root test successfully shows convergence as all sub-limits are less than 1.

Step-by-step solution

Identify the general term of the series

The given series is: \[ x + y + x^{2} + y^{2} + x^{3} + y^{3} + \cdots \] The odd terms are powers of x and the even terms are powers of y. So, the general term can be expressed as: \[ a_{n} = \begin{cases} x^{(n+1)/2} & \text{if } n \text{ is odd} \ y^{n/2} & \text{if } n \text{ is even} \end{cases} \]

D'Alembert Ratio Test

The D'Alembert ratio test requires finding the limit: \[ L = \frac{a_{n+1}}{a_{n}} \] Considering even and odd terms separately: \[ L_{\text{odd}} = \frac{y^{(n+1)/2}}{x^{(n+1)/2}} = \frac{y}{x} > 1 \] \[ L_{\text{even}} = \frac{x^{(n+2)/2}}{y^{n/2}} = x(\frac{x}{y}) \rightarrow 0 \text{ as } x \rightarrow y \] Since the limit does not consistently give values less than 1, the D'Alembert ratio test is inconclusive.

Cauchy Root Test

The Cauchy root test requires finding the limit: \[ L = \root{n}\frown{a_n} \] For odd and even terms, this becomes: \[ L_{\text{odd}} = \root n \frown{ x^{(n+1)/2}} = x^{(n+1)/(2n)} \rightarrow x^{1/2} \text{ as } n \rightarrow \infty \rightarrow \] \[ L_{\text{even}} = \root n \frown{ y^{n/2} } = y^{1/2} \text{ as } n \rightarrow \infty \] Since both limits are less than 1 for 0 < x < y < 1, the Cauchy root test confirms the convergence of the series.

Key concepts

D'Alembert Ratio Test

The D'Alembert ratio test is a method used to determine the convergence or divergence of an infinite series. To apply this test, one must compute the limit of the ratio of successive terms of the series. Specifically, for a series \(\frac{a_{n+1}}{a_n}\), the rule states:

• If the limit \(L = \frac{a_{n+1}}{a_n}<\)1 , the series converges absolutely.
• If \(L >1\), the series diverges.
• If \(L =1\), the test is inconclusive.

In our example, considering even and odd terms separately, we get distinct results:

• For odd terms: \( L_{\text{odd}} = \frac{y}{x} >1 \) as per the given condition \( 0 < x < y < 1\).
• For even terms: \( L_{\text{even}} = x (\frac{x}{y}) \rightarrow 0 \) : as \( x \rightarrow y \).

Due to these mixed results, the D'Alembert ratio test fails to give a conclusive answer on the series' convergence.

Cauchy Root Test

The Cauchy root test is another method used to determine the convergence of an infinite series. This test involves taking the n-th root of the absolute value of the terms of the series. Specifically, for a series \(a_n\), the rule states:

• If \(L = \lim_{n\to\infty} \root{n}\frown{ |a_n| } < 1 \) , the series converges absolutely.
• If \(L > 1 \), the series diverges.
• If \(L = 1 \), the test is again inconclusive.

In our example, considering odd and even terms separately, we calculate:

• For odd terms: \(L_{\text{odd}} = \lim_{n\to\infty} \root{n}\frown{ x^{(n+1)/2}} = x^{1/2} \).
• For even terms: \(L_{\text{even}} = \lim_{n\to\infty} \root{n}\frown{ y^{n/2}} = y^{1/2} \).

Since both limits are less than 1 for \( 0 < x < y < 1 \), the Cauchy root test confirms that the series converges.

General Term Identification

Identifying the general term of a series is a crucial step in applying convergence tests. It helps determine the behavior of the series as n approaches infinity. In our given exercise, the series is:

• \( x + y + x^{2} + y^{2} + x^{3} + y^{3} + \cdots \)

The series alternates between powers of \(x\) and \(y\). By observing this pattern, we can deduce the general term:

• For odd n: \(a_n = x^{(n+1)/2} \).
• For even n: \(a_n = y^{n/2} \).

With this information, we can properly apply the convergence tests like the D'Alembert ratio test and the Cauchy root test. This identification step ensures we approach the problem systematically and accurately.