08 More Convergence Criteria, Dirichlet's Theorem, Ratio, Root Tests, and Power Series

In this lecture, we will expand our toolbox of convergence criteria for series. We will introduce the powerful Ratio and Root tests, which are particularly useful for determining the convergence of series where terms involve factorials or powers. Finally, we will begin our exploration of power series, a fundamental concept in analysis.

Recap of Known Convergence Criteria

Before introducing new tests, let’s recall the convergence criteria we have already discussed:

• Leibniz Criterion (Alternating Series Test): Applicable to alternating series that satisfy specific conditions (decreasing term magnitudes, limit to zero).
• Cauchy Criterion: A fundamental criterion that directly tests the behavior of partial sums, allowing us to determine convergence without knowing the limit.
• Comparison Test (Vergleichssatz): Useful for comparing a given series with another series whose convergence behavior is known.

From the previous lecture, we remember the Corollary of the Comparison Test (Vergleichssatz):

Corollary (Comparison Test - Revisited): Let ${\sum }_{k=1}^{\infty }{a}_k$ and ${\sum }_{k=1}^{\infty }{b}_k$ be series with $0\le a_k\le b_k$ for all $k\ge K$ for some $K\ge 1$. Then:

1. If ${\sum }_{k=1}^{\infty }{b}_k$ converges, then ${\sum }_{k=1}^{\infty }{a}_k$ converges.
2. If ${\sum }_{k=1}^{\infty }{a}_k$ diverges, then ${\sum }_{k=1}^{\infty }{b}_k$ diverges.

This comparison test is a valuable tool, but it requires finding suitable series to compare with. The Ratio and Root tests offer more direct methods, often easier to apply in many situations.

Dirichlet’s Theorem

But first, let us prove Dirichlet’s Theorem.

Rearrangements of Absolutely Convergent Series

As we learned from the Riemann Rearrangement Theorem, conditionally convergent series exhibit highly unstable behavior under rearrangements. However, absolutely convergent series are remarkably well-behaved in this regard. Dirichlet’s Theorem formalizes this stability.

Theorem (Dirichlet’s Theorem): If ${\sum }_{k=1}^{\infty }{a}_k$ is absolutely convergent, then every rearrangement ${\sum }_{k=1}^{\infty }{a}_k^{\prime }$ of the series is also absolutely convergent, and every rearrangement has the same sum.

This is a powerful result! It tells us that for absolutely convergent series, we can rearrange the terms in any way we like without affecting the sum or losing absolute convergence. This property is not shared by conditionally convergent series, making absolute convergence a highly desirable property.

Proof of Dirichlet’s Theorem

Let ${\sum }_{k=1}^{\infty }{a}_k$ be an absolutely convergent series. Let ${\sum }_{k=1}^{\infty }{a}_k^{\prime }$ be a rearrangement of ${\sum }_{k=1}^{\infty }{a}_k$, where $a_k^{\prime }=a_{\sigma (k)}$ and $\sigma :{\mathbb{N}}^{\ast }\to {\mathbb{N}}^{\ast }$ is a bijection.

We need to show two things:

1. Absolute convergence
2. Equality

(1) Absolute Convergence of the Rearrangement ${\sum }_{k=1}^{\infty }{a}_k^{\prime }$

Since ${\sum }_{k=1}^{\infty }|a_k|$ converges, let $T={\sum }_{k=1}^{\infty }|a_k|$ be its sum (a finite value as it converges).

Consider the partial sums of the rearranged series of absolute values: ${\sum }_{k=1}^n|a_k^{\prime }|={\sum }_{k=1}^n|a_{\sigma (k)}|$.

Let $N=\max \{\sigma (1),\sigma (2),\dots ,\sigma (n)\}$. Then the set of indices $\{\sigma (1),\sigma (2),\dots ,\sigma (n)\}$ is a subset of $\{1,2,\dots ,N\}$. Therefore,

${\sum }_{k=1}^n|a_k^{\prime }|={\sum }_{k=1}^n|a_{\sigma (k)}|\le {\sum }_{k=1}^N|a_k|\le {\sum }_{k=1}^{\infty }|a_k|=T$

Thus, the partial sums ${\sum }_{k=1}^n|a_k^{\prime }|$ are bounded above by $T$. Since ${\sum }_{k=1}^{\infty }|a_k^{\prime }|$ is a series with non-negative terms and its partial sums are bounded above, by the criterion for convergence of non-negative series, ${\sum }_{k=1}^{\infty }|a_k^{\prime }|$ converges. This shows that the rearrangement ${\sum }_{k=1}^{\infty }{a}_k^{\prime }$ is absolutely convergent.

(2) Equality of Sums: ${\sum }_{k=1}^{\infty }{a}_k^{\prime }={\sum }_{k=1}^{\infty }{a}_k$

Let $S={\sum }_{k=1}^{\infty }{a}_k$. We want to show that a rearrangement ${\sum }_{k=1}^{\infty }{a}_{\sigma (k)}^{\prime }$ also converges to $S$. Let $S_m^{\prime }={\sum }_{k=1}^m{a}_{\sigma (k)}^{\prime }$ be the partial sums of the rearranged series. We need to show that ${\lim }_{m\to \infty }{S}_m^{\prime }=S$.

Given $ϵ>0$. Since ${\sum }_{k=1}^{\infty }|a_k|$ converges, by the Cauchy Criterion, there exists an $N\in \mathbb{N}$ such that for all $n>N$:

${\sum }_{k=n+1}^{\infty }|a_k|<ϵ$

Choose $M$ large enough so that the set of indices $\{\sigma (1),\sigma (2),\dots ,\sigma (M)\}$ contains all indices $\{1,2,\dots ,N\}$. This is possible because $\sigma$ is a bijection.

Consider the difference between the partial sum of the rearranged series $S_M^{\prime }$ and the sum of the original series $S$:

$|S_M^{\prime }-S|=\left|{\sum }_{k=1}^M{a}_{\sigma (k)}-{\sum }_{k=1}^{\infty }{a}_k\right|=\left|-{\sum }_{k\notin \{\sigma (1),...,\sigma (M)\}}{a}_k\right|=\left|{\sum }_{k\notin \{\sigma (1),...,\sigma (M)\}}{a}_k\right|$

Since $\{\sigma (1),\dots ,\sigma (M)\}\supseteq \{1,2,\dots ,N\}$, any index $k\notin \{\sigma (1),\dots ,\sigma (M)\}$ must be greater than $N$. Therefore, the sum ${\sum }_{k\notin \{\sigma (1),...,\sigma (M)\}}{a}_k$ is a sum over a subset of indices that are all greater than $N$.

We can bound the absolute value of this sum by summing the absolute values of the terms and using the tail estimate from the Cauchy Criterion:

$|S_M^{\prime }-S|\le {\sum }_{k\notin \{\sigma (1),...,\sigma (M)\}}|a_k|\le {\sum }_{k=N+1}^{\infty }|a_k|<ϵ$

Thus, for sufficiently large $M$, $|S_M^{\prime }-S|<ϵ$. This shows that ${\lim }_{M\to \infty }{S}_M^{\prime }=S$, which means ${\sum }_{k=1}^{\infty }{a}_k^{\prime }=S={\sum }_{k=1}^{\infty }{a}_k$.

Therefore, every rearrangement of an absolutely convergent series converges to the same sum.

Tests

Now let us continue with some other tests to check for convergence or divergence.

The Ratio Test (Quotientenkriterium)

Statement of the Ratio Test

The Ratio Test examines the ratio of consecutive terms in a series to determine its convergence.

Theorem (Ratio Test): Let $(a_n{)}_{n\ge 1}$ be a sequence of real or complex numbers with $a_n\ne 0$ for all $n\ge 1$.

1. Convergence: If ${\mathrm{limsup}}_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|<1$, then the series ${\sum }_{k=1}^{\infty }{a}_k$ converges absolutely.

2. Divergence: If ${\mathrm{liminf}}_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|>1$, then the series ${\sum }_{k=1}^{\infty }{a}_k$ diverges.

Note: If ${\mathrm{limsup}}_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|=1$, the Ratio Test is inconclusive; we cannot determine convergence or divergence based on this test alone.

Intuition Behind the Ratio Test

The Ratio Test is based on the idea of comparing the series to a geometric series. If the ratio of consecutive terms is consistently less than 1 in absolute value, it suggests that the terms are decreasing rapidly enough for the series to converge, similar to a convergent geometric series. Conversely, if the ratio is consistently greater than 1, the terms are not decreasing sufficiently, and the series will likely diverge.

Proof of the Ratio Test (Convergence Case)

For divergence…

For convergence…

Let $L={\mathrm{limsup}}_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|<1$. Choose a value $q$ such that $L<q<1$.

By the property of the limit superior (from a previous lecture on limsup properties), for $ϵ=q-L>0$, there exists an index $K\in \mathbb{N}$ such that for all $k\ge K$,

$\left|\frac{a_{k+1}}{a_k}\right|<{\mathrm{limsup}}_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|+ϵ=L+(q-L)=q<1$

Thus, for all $k\ge K$, we have $\left|\frac{a_{k+1}}{a_k}\right|\le q$, which implies $|a_{k+1}|\le q\cdot |a_k|$.

Applying this inequality repeatedly for $n>K$:

Let $C=\frac{|a_K|}{q^K}$ (which is a constant). Then for $n>K$, $|a_n|\le C\cdot q^n$.

Since $0<q<1$, the geometric series ${\sum }_{n=K+1}^{\infty }C\cdot q^n=C{\sum }_{n=K+1}^{\infty }{q}^n$ converges. By the Comparison Test, the series ${\sum }_{n=K+1}^{\infty }|a_n|$ also converges. Therefore, ${\sum }_{k=1}^{\infty }|a_k|$ converges, and the series ${\sum }_{k=1}^{\infty }{a}_k$ is absolutely convergent.

Example: Exponential Function Series

Consider the series related to the exponential function: ${\sum }_{k=0}^{\infty }\frac{z^k}{k!}$ for a complex number $z\ne 0$. Let $a_k=\frac{z^k}{k!}$. Then:

$\left|\frac{a_{k+1}}{a_k}\right|=\left|\frac{z^{k+1}/(k+1)!}{z^k/k!}\right|=\left|\frac{z^{k+1}}{(k+1)!}\cdot \frac{k!}{z^k}\right|=\left|\frac{z}{k+1}\right|=\frac{|z|}{k+1}$

Now, we compute the limit superior:

${\mathrm{limsup}}_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|={\mathrm{limsup}}_{n\to \infty }\frac{|z|}{n+1}=0$

Since $0<1$, by the Ratio Test, the series ${\sum }_{k=0}^{\infty }\frac{z^k}{k!}$ converges absolutely for any $z\ne 0$. It also converges for $z=0$ trivially. This series defines the exponential function, $\exp (z)={\sum }_{k=0}^{\infty }\frac{z^k}{k!}$.

Limitations of the Ratio Test

The Ratio Test is inconclusive when ${\mathrm{limsup}}_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|=1$. In such cases, we need to use other convergence tests. For example, for p-series $\sum \frac{1}{n^p}$, the ratio test limit is always 1, regardless of whether the series converges ($p>1$) or diverges ($p\le 1$).

The Root Test (Wurzelkriterium)

Statement of the Root Test

The Root Test examines the $n$-th root of the absolute value of the terms to determine convergence.

Theorem (Root Test): Let $(a_n{)}_{n\ge 1}$ be a sequence of real or complex numbers.

1. Convergence: If ${\mathrm{limsup}}_{n\to \infty }\sqrt[n]{|a_n|}<1$, then the series ${\sum }_{k=1}^{\infty }{a}_k$ converges absolutely.

2. Divergence: If ${\mathrm{limsup}}_{n\to \infty }\sqrt[n]{|a_n|}>1$, then the series ${\sum }_{k=1}^{\infty }{a}_k$ diverges.

Note: Similar to the Ratio Test, if ${\mathrm{limsup}}_{n\to \infty }\sqrt[n]{|a_n|}=1$, the Root Test is inconclusive.

Intuition Behind the Root Test

The Root Test also compares the series to a geometric series. If the $n$-th root of $|a_n|$ is consistently less than 1, it suggests that $|a_n|$ is decaying faster than a geometric sequence with ratio less than 1, leading to convergence.

Proof of the Root Test (Convergence Case)

For divergence…

For convergence…

Let $L={\mathrm{limsup}}_{n\to \infty }\sqrt[n]{|a_n|}<1$. Choose a value $q$ such that $L<q<1$.

By the property of the limit superior, for $ϵ=q-L>0$, there exists an index $K\in \mathbb{N}$ such that for all $k\ge K$,

$\sqrt[k]{|a_k|}<{\mathrm{limsup}}_{n\to \infty }\sqrt[n]{|a_n|}+ϵ=L+(q-L)=q<1$

Thus, for all $k\ge K$, we have $\sqrt[k]{|a_k|}<q$, which implies $|a_k|<q^k$.

Since $0<q<1$, the geometric series ${\sum }_{k=K}^{\infty }{q}^k$ converges. By the Comparison Test, the series ${\sum }_{k=K}^{\infty }|a_k|$ also converges. Therefore, ${\sum }_{k=1}^{\infty }|a_k|$ converges, and the series ${\sum }_{k=1}^{\infty }{a}_k$ is absolutely convergent.

Example: Applying the Root Test

Consider the series ${\sum }_{k=1}^{\infty }(\frac{k+(-1{)}^k}{2k}{)}^k$. Let $a_k=(\frac{k+(-1{)}^k}{2k}{)}^k$. Then:

$\sqrt[k]{|a_k|}=\sqrt[k]{\left|(\frac{k+(-1{)}^k}{2k}{)}^k\right|}=\left|\frac{k+(-1{)}^k}{2k}\right|=\frac{|k+(-1{)}^k|}{2k}$

For large $k$, $(-1{)}^k$ is either 1 or -1, so $k+(-1{)}^k$ is close to $k$.

${\lim }_{k\to \infty }\sqrt[k]{|a_k|}={\lim }_{k\to \infty }\frac{|k+(-1{)}^k|}{2k}={\lim }_{k\to \infty }\frac{k+(-1{)}^k}{2k}={\lim }_{k\to \infty }\frac{1+\frac{(-1{)}^k}{k}}{2}=\frac{1}{2}$

Since ${\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|a_k|}=\frac{1}{2}<1$, by the Root Test, the series ${\sum }_{k=1}^{\infty }(\frac{k+(-1{)}^k}{2k}{)}^k$ converges absolutely.

Comparing Ratio and Root Tests

• The Root Test is generally more powerful than the Ratio Test. If the Ratio Test determines convergence or divergence, the Root Test will also lead to the same conclusion. However, there are series where the Root Test can determine convergence or divergence, but the Ratio Test is inconclusive.
• In practice, the Ratio Test is often easier to apply when terms involve factorials or simple exponential terms. The Root Test is more suitable when terms involve $k$-th powers, as the $k$-th root simplifies the expression.

For the example above, applying the Ratio Test is more complex:

$\left|\frac{a_{k+1}}{a_k}\right|=\frac{(\frac{k+1+(-1{)}^{k+1}}{2(k+1)}{)}^{k+1}}{(\frac{k+(-1{)}^k}{2k}{)}^k}=(\frac{k+1+(-1{)}^{k+1}}{2(k+1)}{)}^{k+1}\cdot (\frac{2k}{k+(-1{)}^k}{)}^k$

Analyzing the limit superior of this ratio is more involved than calculating the limit superior of the $k$-th root.

Power Series (Potenzreihen)

Power series are a cornerstone of analysis, providing a way to represent functions as infinite polynomials. They are incredibly versatile, enabling us to study familiar functions like exponentials and trigonometric functions in a new light, and to define entirely new functions.

Definition: The Infinite Polynomial

A power series centered at $0$ is an infinite series of the form:

${\sum }_{k=0}^{\infty }{c}_k{z}^k=c_0+c_{1}z+c_2{z}^2+c_3{z}^3+\dots$

where:

• $(c_k{)}_{k\ge 0}$ is a sequence of coefficients (real or complex numbers).
• $z$ is a complex variable.

For each choice of coefficients $(c_k)$, we get a different power series, and for each value of $z$, the series may converge or diverge, defining a function in its region of convergence.

Intuition: Why a Radius of Convergence?

Imagine testing the convergence of a power series for different values of $z$. Intuitively, for small values of $|z|$, the terms $c_k{z}^k$ become smaller faster as $k$ increases (especially if the coefficients $c_k$ don’t grow too rapidly). This suggests convergence near $z=0$. Conversely, for large $|z|$, the terms might grow, leading to divergence.

The remarkable fact is that for every power series, there exists a radius of convergence, $\rho$, that precisely delineates the region of convergence as a disk centered at the origin in the complex plane.

Definition: Convergence Radius

The radius of convergence $\rho$ of a power series ${\sum }_{k=0}^{\infty }{c}_k{z}^k$ is given by:

$\rho =\frac{1}{{\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}}$

with the following conventions:

• If ${\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}=0$, then $\rho =\infty$.
• If ${\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}=\infty$, then $\rho =0$.

This formula, derived directly from the Root Test, elegantly captures the boundary of convergence.

Theorem: Convergence Behavior Based on Radius

Theorem (Convergence of Power Series): For a power series ${\sum }_{k=0}^{\infty }{c}_k{z}^k$ with convergence radius $\rho$:

1. Absolute Convergence Inside the Disk: For all $z\in \mathbb{C}$ with $|z|<\rho$, the power series ${\sum }_{k=0}^{\infty }{c}_k{z}^k$ converges absolutely.

2. Divergence Outside the Disk: For all $z\in \mathbb{C}$ with $|z|>\rho$, the power series ${\sum }_{k=0}^{\infty }{c}_k{z}^k$ diverges.

3. On the Circle $|z|=\rho$: The convergence behavior on the circle $|z|=\rho$ is more intricate and requires separate investigation for each specific power series and point $z$ on the circle. The series may converge for some points and diverge for others on the circle.

Proof

We will use the Root Test to prove the convergence and divergence properties. Let $a_k=c_k{z}^k$. We need to analyze ${\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|a_k|}={\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k{z}^k|}={\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}|z|=|z|{\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}$.

Let $L={\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}$. Then $\rho =\frac{1}{L}$ (with the conventions for $L=0$ and $L=\infty$). We are examining the value $|z|L$.

(1) Convergence for $|z|<\rho$:

If $|z|<\rho$, then $|z|<\frac{1}{L}$, so $|z|L<1$. Applying the Root Test: ${\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|a_k|}=|z|L<1$. By the Root Test, the series ${\sum }_{k=0}^{\infty }{c}_k{z}^k$ converges absolutely.

(2) Divergence for $|z|>\rho$:

If $|z|>\rho$, then $|z|>\frac{1}{L}$, so $|z|L>1$. Applying the Root Test: ${\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|a_k|}=|z|L>1$. By the Root Test, the series ${\sum }_{k=0}^{\infty }{c}_k{z}^k$ diverges.

(3) Inconclusive for $|z|=\rho$:

If $|z|=\rho$, then $|z|=\frac{1}{L}$, so $|z|L=1$. In this case, ${\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|a_k|}=|z|L=1$. The Root Test is inconclusive, and we need to investigate convergence on the circle $|z|=\rho$ using other methods for each specific power series.

Example: Finding the Radius of Convergence

Consider the power series ${\sum }_{k=1}^{\infty }\frac{k}{2^k}{z}^k$. Here, $c_k=\frac{k}{2^k}$.

${\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}={\mathrm{limsup}}_{k\to \infty }\sqrt[k]{\frac{k}{2^k}}=\frac{1}{2}{\mathrm{limsup}}_{k\to \infty }\sqrt[k]{k}=\frac{1}{2}\cdot 1=\frac{1}{2}$

The convergence radius is $\rho =\frac{1}{1/2}=2$. The series converges absolutely for $|z|<2$ and diverges for $|z|>2$.

Power series provide a bridge between sequences, series, and functions. Understanding their convergence properties, especially through the concept of the radius of convergence, is fundamental for working with functions defined by infinite sums in complex and real analysis.

This lecture expanded our knowledge of convergence criteria by introducing the Ratio and Root tests, powerful tools for determining the convergence of series, particularly in cases involving ratios or roots of terms. We also introduced the concept of power series and the convergence radius, setting the stage for further exploration of functions defined by series.