Question

Examples: Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) A series that converges absolutely.

(b) A series that converges conditionally.

(c) A series \[\sum _{k=1}^{\infty}a_{k}\] such that \[\lim_{k \to \infty} \left | \frac{a_{k+1}}{a_{k}} \right | =1\] but the series is absolutely convergent.

Short answer

(a) The series which converges absolutely is Taylor expansion of the natural logarithmic function

\[ln(x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+......\]

(b) The series which converges conditionally is

\[\frac{(-1)^{n}}{\sqrt{n}}=-\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......\]

(c) It is not possible for a series to have a convergence limit of 1 for the ratio of consecutive terms i.e

\[\lim_{k \to \infty} \left | \frac{a_{k+1}}{a_{k}} \right | =1\]

and be absolutely convergent at the same time.

Step-by-step solution

Step 1. Part (a)

The example of the series that converges absolutely would be the Taylor expansion of the natural logarithmic function:

\[ln(x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+......\]

This series converges for all values of x such that |x| < 1, and it converges absolutely because the absolute value of each term is less than x.

The radius of convergence of a Taylor series is the largest interval around the expansion point in which the series converges. The radius of convergence of the Taylor series for the natural logarithm function is 1, and the series converges absolutely within this interval.

Step 2. Part(b)

The example of the series that converges conditionally is the series:

\[\frac{(-1)^{n}}{\sqrt{n}}=-\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......\]

This series converges to approximately 2, but it converges conditionally because it does not satisfy the absolute convergence test (the absolute value of each term is not less than 1), but it does satisfy the alternating series test (the terms of the series alternate in sign and decrease in absolute value).

Step 3. Part (c)

If the series has the limit of the ratio of the consecutive terms

\[\lim_{k \to \infty} \left | \frac{a_{k+1}}{a_{k}} \right | =1\]

If the convergence limit is 1, then the ratio of consecutive terms will approach 1 as k increases. This means that the terms of the series will be approximately equal in size, and the sum of the series will not converge.

On the other hand, if the convergence limit is less than 1, then the ratio of consecutive terms will approach a value less than 1 as k increases. This means that the terms of the series will decrease in size, and the sum of the series will converge.

Therefore, it is not possible for a series to have a convergence limit of 1 for the ratio of consecutive terms and be absolutely convergent at the same time.