# Power series

## 1The idea of radius of convergence

Definition 1.1

A power series is a series of the form $∑ n = 0 ∞ a n x n$, where $a n ∈ ℝ$ for each $n ∈ ℕ$ and $x ∈ ℝ$.

Such series are often used to define functions, such as

$f ( x ) = ∑ n = 0 ∞ a n x n .$

Obviously, if we are going to define new functions in this way, it is important to know when such a series converges. As indicated in the last equation, we generally think of the numbers $a n$ as fixed numbers (or fixed coefficients) and the $x$ as a variable.

Power series are also used to define functions on the complex numbers too:

$f ( z ) = ∑ n = 0 ∞ a n z n$

for $z ∈ ℂ$. Although these notes are primarily about real numbers and real analysis, and our proofs are presented for the reals only, the area of power series is one important example where what we shall say also works equally well, and to great effect, for the complex numbers too.

The main result about power series is that each power series has a radius of convergence, $R ≥ 0$, which is a real number such that:

• if $x < R$ then $∑ n = 0 ∞ a n x n$ converges; and
• if $x > R$ then $∑ n = 0 ∞ a n x n$ diverges.

This result, i.e., the existence of such a radius of convergence, in fact works equally well for the complexes as for the reals. The terminology radius arises from thinking about the set of complex numbers $z$ with $z < R$ in the Argand diagram: this set is a circular disc with certre $0$ and radius $R$.

The radius of convergence, $R$ could turn out to be any number between $0$ and $∞$, including both extremes $0$ and $∞$. Of course if $R = ∞$ that means that the power series converges for all possible values of $x$. If $R = 0$ it means that the power series diverges for all possible values of $x$, except $x = 0$ itself. (It is easy to see that all power series of the form $∑ n = 0 ∞ a n x n$ converge for $x = 0$ as there is at most one nonzero term in the series, the one for $a 0$.)

One further point to bear in mind is that the theorem on radius of convergence says nothing about what actually happens on the perimeter of this circular disc, i.e., when $z = R$. In fact, for some power series, the series converges everywhere on this boundary, and for others it converges nowhere on the boundary. For yet others, it converges at some points but not at others in a rather complicated way. This point is related to the ratio test, where the limit of the ratio is $1$: in such cases, the situation may be highly complicated and a simple test cannot give the answer.

## 2Proof of the radius of convergence theorem

We shall consider a power series $∑ n = 0 ∞ a n x n$ and prove it has a radius of convergence, $R$, as defined above. The main work involved takes place in the following theorem.

Theorem 2.1

Suppose $x 0 ∈ ℝ$ and $∑ n = 0 ∞ a n x 0 n$ converges, and let $x ∈ ℝ$ with $x < x 0$. Then $∑ n = 0 ∞ a n x n$ converges absolutely.

Proof

We have $a n x 0 n → 0$ as $n → 0$, by the null sequence test, so

So, as $x < x 0$, we have for all $n ≥ N 0$

$a n x n = a n x 0 n ⋅ x x 0 n < 1 2 x x 0 n ,$

so $∑ n = 0 ∞ a n x n$ converges by the comparison test, using comparison with the series

$∑ n = 0 ∞ x x 0 n$

which converges as $x x 0 < 1$.

Surprisingly, perhaps, this does all the work we need, and to get the radius of convergence theorem we just need to combine the last theorem with an old result about completeness of the reals.

For all power series $∑ n = 0 ∞ a n x n$ there is a nonnegative radius of convergence $R ∈ ℝ ∪ ∞$ such that the power series converges absolutely for all $x$ with $x < R$ and the power series diverges for all $x$ with $x > R$.

Proof

Let $A$ be the set

There are two cases depending on whether or not $A$ is bounded.

Subproof

Case 1: $A$ is unbounded. We show we can take $R = ∞$ in this case.

Subproof

Let $x ∈ ℝ$ be arbitrary.

Then because $A$ is unbounded there is $r ∈ A$ with $x < r$ and hence there is $x 0 ∈ ℝ$ with $x 0 = r > x$ and $∑ n = 0 ∞ a n x 0 n$ converging. Thus by the previous result $∑ n = 0 ∞ a n x n$ converges absolutely.

Therefore $∑ n = 0 ∞ a n x n$ converges absolutely for every $x ∈ ℝ$, as required.

On the other hand,

Subproof

Case 2: $A$ is bounded.

In this case, $A$ is a bounded nonempty set (nonempty because it contains $0$) and so by the supremum form of the completeness of reals there is a least upper bound $R = sup A$ of $A$. We have to show $∑ n = 0 ∞ a n x n$ converges for $x < R$ and diverges for $x > R$.

Subproof

Let $x ∈ ℝ$ with $x < R$ be arbitrary.

Then since $R$ is the least upper bound of $A$ and $x$ is smaller, $x$ is not an upper bound of $A$ so there is $r ∈ A$ with $x < r$ and hence there is $x 0 ∈ ℝ$ with $x 0 = r > x$ and $∑ n = 0 ∞ a n x 0 n$ converging. Thus by the previous result $∑ n = 0 ∞ a n x n$ converges absolutely.

So . Also,

Subproof

Let $x ∈ ℝ$ with $x > R$ be arbitrary.

Then if $∑ n = 0 ∞ a n x n$ converges we would have $x ∈ A$. But this is impossible as $R$ is an upper bound of $A$ and $R < x$. Therefore $∑ n = 0 ∞ a n x n$ does not converge.

So .

## 3Examples

Example 3.1

The series $∑ n = 0 ∞ x n$ has radius of convergence $1$. It converges absolutely if $x < 1$, and diverges for all other $x$ including $1$ and $-1$.

Proof

The ratio of consecutive terms of $∑ n = 0 ∞ x n$ is $x n + 1 x n = x$. This is less than $1$ if $x < 1$, in which case the series converges absolutely by the ratio test. It is greater than one if $x > 1$, in which case the series diverges by the ratio test. Therefore the radius of convergence is $1$.

On the radius of convergence we have $x = 1$ or $x = -1$ and the series is either $∑ n = 0 ∞ 1 n$ or $∑ n = 0 ∞ -1 n$. Both these diverge by the null sequence test.

Example 3.2

The series $∑ n = 0 ∞ x n n$ has radius of convergence $1$. It converges absolutely if $x < 1$, converges conditionally when $x = -1$ and diverges for all other $x$ including $1$.

Proof

The ratio of consecutive terms of $∑ n = 0 ∞ x n n$ is $n x n + 1 = x 1 + n -1$ which converges to $x$. Therefore, using the ratio test as in the last example, the radius of convergence is $1$ and the series converges absolutely when $x < 1$ and diverges when $x > 1$. For $x = 1$ our series is the harmonic series, which diverges, and for $x = -1$ our series converges by the alternating series test, but does not converge absolutely as the series of absolute values of terms is again the harmonic series.

Example 3.3

The series $∑ n = 0 ∞ x n n 2$ has radius of convergence $1$. It converges absolutely if $x ≤ 1$, and diverges for all other $x$.

Proof

Again, the ratio of consecutive terms of $∑ n = 0 ∞ x n n$ is $n 2 x ( n + 1 ) 2 = x 1 + 2 n -1 + n -2$ which converges to $x$. Therefore, by the ratio test again, the radius of convergence is $1$ and the series converges absolutely when $x < 1$ and diverges when $x > 1$. For $x = 1$ our series is $∑ n = 0 ∞ n -2$ which converges and is made up of positive terms only, hence converges absolutely. For $x = -1$ it also converges absolutely as the absolute value of the terms in this series are the same terms as in the case $x = 1$.

Example 3.4

The series $∑ n = 0 ∞ x n n !$ has radius of convergence $∞$. It converges absolutely for all $x ∈ ℝ$.

Proof

The ratio of consecutive terms of $∑ n = 0 ∞ x n n !$ is $x n + 1 → 0$ as $n → ∞$, so by the ratio test the series converges absolutely for all $x ∈ ℝ$.

Example 3.5

The series $∑ n = 0 ∞ n ! x n$ has radius of convergence $0$. It converges for $x = 0$ and diverges for all $x ≠ 0$.

Proof

For $x ≠ 0$ the sequence $n ! x n$ is not null. To see this, let $y = x -1$ and let $k ∈ ℕ$ be any odd positive integer with $k > y$. Then $1 k k 2 k ≥ 1$, $2 k k 2 - 1 k ≥ 1$, $3 k k 2 - 2 k ≥ 1$, and so on, hence $k 2 ! ≥ k k 2$. This is the base case of an induction argument that shows that $n ! ≥ k n$ for $n ≥ k 2$. (The induction step is easy as $n > k$ for such $n$.) Since $k > y = x -1$ it follows that $n ! x n ≥ 1$ for $n ≥ k 2$ and $n ! x n$ is not a null sequence.

Hence from the null sequence test that the series $∑ n = 0 ∞ n ! x n$ does not converge absolutely for any $x ≠ 0$. From this it follows by the theorem on power series given above that $∑ n = 0 ∞ n ! x n$ does not converge at all for any $x ≠ 0$. For if this series did converge with $x ≠ 0$ then $∑ n = 0 ∞ n ! x 2 n$ would also converge.

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