44. Alternating Series¹ and Conditional Convergence

Dated: 30-10-2024

Alternating Series¹

The series¹ of the form

\[\sum_{k=1}^{\infty}(-1)^{k+1}a_{k}=a_{1}-a_{2}+a_{3}-a_{4}+\ldots\]

\[\sum_{k=1}^{\infty}(-1)^{k}a_{k}=a_{1}-a_{2}+a_{3}-a_{4}+\ldots\]

Are called alternating series.

Alternating Series Test

The alternating series converges if the following 2 conditions are satisfied

1. \[a_1 > a_2 > \ldots > a_i > \ldots\]
2. \[\lim_{i \to +\infty} a_i = 0\]

Theorem

If an alternating series satisfies the alternating series test and the sum² is approximated by \(n^{th}\) partial sum² \(S_n\), thereby resulting in an error of \(S - S_n\) then

\[\lvert S - S_n \rvert < a_{n + 1}\]

The sign of the error is same as the coefficient of \(a_{n+1}\).

Example

Assume an alternating series

\[\sum_{i = 1}^{\infty} (-1)^{i + 1} \frac 1 i\]

This alternating series satisfies the conditions of alternating series test which implies that it converges which further implies that it has a sum² \(S\) which must lie between 2 successive partial sums.²

\[S_{7}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}=\frac{319}{420}\]

\[S_{8}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}=\frac{533}{840}\]

\[S_7 < S < S_8\]

We can arbitrarily choose \(S = \ln (2)\) because it satisfies the inequality.³

\[\lvert \ln 2-S_{7}\rvert = \lvert \ln 2-\frac{319}{420}\rvert <a_{8}=\frac{1}{8}\]

\[\lvert \ln 2-S_{8}\rvert = \lvert \ln 2-\frac{533}{840}\rvert<a_{9}=\frac{1}{9}\]

Absolute Convergence

A series¹

\[\sum_{i=1}^{\infty}u_{i}=u_{1}+u_{2}+ \ldots +u_{i}+ \ldots\]

is said to converge absolutely if the series¹ of its absolute values⁴ converges.

\[\sum_{i=1}^{\infty}\lvert u_{i}\rvert=\lvert u_{1}\rvert+\lvert u_{2}\rvert+\ldots+\lvert u_{i}\rvert+\ldots\]

Conditional Convergence

Imagine a series¹ like

\[\sum_{i = 1}^{\infty} (-1)^{i + 1} \frac 1 i = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+(-1)^{i+1}\frac{1}{i}+\ldots\]

Which converges but the following series¹ diverges

\[\sum_{i = 1}^{\infty}\lvert (-1)^{i + 1} \frac 1 i \rvert = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+(-1)^{i+1}\frac{1}{i}+\ldots\]

Hence, it is not absolutely convergent.
Such a series¹ is called conditionally convergent.

Ratio Test for Absolute Convergence

Suppose that \(\sum u_i\) be a series¹ with non zero terms and

\[p=\lim_{i\rightarrow\infty}\frac{\lvert u_{i+1}\rvert}{\lvert u_{i}\rvert}\]

Then if

1. \(p < 1\), then \(\sum u_i\) converges absolutely.
2. \(p > 1\), then \(\sum u_i\) diverges.
3. \(p = 1\), no conclusion can be drawn about convergence or absolute convergence.

Power Series in \(x\)

If \(c_0, c_1\) etc are constants and \(x\) is a variable then the series¹ of the form

\[\sum_{i=0}^{\infty}c_{i}x^{i}=c_{0}+c_{1}x+c_{2}x^{2}+\ldots+c_{i}x^{i}+\ldots\]

is called power series.

Theorem

For power series, exactly one of the following is true

1. The series¹ converges only for \(x = 0\)
2. The series¹ converges absolutely for all real values of \(x\).
3. The series¹ converges absolutely for all \(x\) in some finite open interval⁵ \((-R, R)\), diverges if \(x < -R\) or \(x > R\) and at \(x = -R, R\), can either
   1. converge
   2. converge absolutely
   3. diverge

Radius and Interval of Convergence

The set of values of \(x\) where the power series converges is always an interval⁵ centered at \(0\).
This interval⁵ is called interval of convergence.
Corresponding to which, the series has radius called radius of convergence.

Example

Find interval of convergence and radius of convergence of the following power series

\[\sum_{i = 1}^{\infty} x ^i\]

Solution

Applying the ratio test

\[p=\lim_{i\rightarrow\infty}\frac{\lvert u_{i+1}\rvert}{\lvert u_{i}\rvert}\]

\[=\lim_{i\rightarrow\infty}\frac{\lvert {x^{i + 1}}\rvert}{\lvert x^{i}\rvert}\]

\[= \lim_{i \to \infty} \lvert x \rvert = \lvert x \rvert\]

So the ratio test implies that the series¹ would converge absolutely if \(p = \lvert x \rvert < 1\) and would diverge if \(p = \lvert x \rvert > 1\)
Since \(\lvert x \rvert = 1\) means either \(x = 1\) or \(x = -1\). Therefore, we have to check the convergence at both points.

\(x = 1\)

\[\sum_{k=0}^{\infty}1^{k}=1+1+1+\ldots\]

\(x = -1\)

\[\sum_{k=0}^{\infty}(-1)^{k}=1-1+1-1+\ldots\]

Both of these diverge. Therefore, the interval of convergence is \((-1, 1)\) and the radius of convergence is \(R = 1\).

Power Series in \(x - a\)

\[\sum_{i=0}^{\infty}c_{i}(x-a)^{i}=c_{0}+c_{1}(x-a)+c_{2}(x-a)^{2}+\ldots+c_{i}(x-a)^{i}+\ldots\]

Theorem

For any power series of the form

\[\sum c_{i}(x-a)^{i}\]

exactly one of the following is true

1. The series¹ converges only for \(x = a\)
2. The series¹ converges absolutely for all real values of \(x\).
3. The series¹ converges absolutely for all \(x\) in some open interval⁵ \((a - R, a + R)\), diverges if \(x < a - R\) or \(x > a + R\) and at \(x = a \pm R\), may
   1. converge
   2. converge absolutely
   3. diverge

The interval of convergence is centered at \(x = a\).

Reference

Read more about notations and symbols.

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1. Read more about series. ↩↩↩↩↩↩↩↩↩↩↩↩↩↩↩↩↩

2. Read more about sum. ↩↩↩↩

3. Read more about inequalities. ↩

4. Read more about absolute values. ↩

5. Read more about intervals. ↩↩↩↩