how to find interval of convergence

ORIGINAL FULL PAGE: Rules for determining interval of convergence of power series
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This article describes a rule, set of rules, or procedure, for determining the set of values for a parameter where a series or integral defined in terms of that parameter converges.
See other rules for determining interval of convergence

1 Statement
2 Explanation
3 Hierarchy of functions
- 3.1 List of functions
- 3.2 Interaction between functions
- 3.3 Remember that all these are functions in
4 Determination of interval of convergence
- 4.1 Superexponential growth and decay
- 4.2 Subexponential growth and decay: degree difference test
- 4.3 Subexponential growth and decay: degree difference test with cautionary use where exponent is a multiple of the index of summation
- 4.4 Subexponential growth and decay: examples where the exponent is a power of the index of summation
- 4.5 Exponential growth or decay: degree difference test on coefficients
- 4.6 Exponential growth or decay: cautionary use where the exponent is a multiple of the summation index
- 4.7 Tricky examples

Statement

Consider a power series of the form:

$\sum_{k=0}^\infty a_kx^k$

The rules here help determine the radius of convergence and interval of convergence for this power series in terms of the manner in which the coefficients $a_k$ grow or decay with $k$ as $k \to \infty$ .

Below is a table with all the cases:

Verbal description of case	Lim sup version for coefficients	Other version for coefficients	Conclusion about radius of convergence	Conclusion about interval of convergence	Examples of functions $k \mapsto a_k$
coefficients decay superexponentially	$\lim \sup_{k \to \infty} \|a_k\|^{1/k} = 0$	[SHOW MORE] For any $0 < \lambda < 1$ , there exists a positive integer $k_0$ such that $\|a_k\| < \lambda^k \ \forall \ k \ge k_0$	$\infty$	all of $\R$	reciprocal of function with doubly exponential growth, reciprocal of exponential of function with polynomial growth, reciprocal of factorial function or factorial-type function.
coefficients grow superexponentially	$\lim \sup_{k \to \infty} \|a_k\|^{1/k} = \infty$		0	$\{ 0 \}$	function with doubly exponential growth, exponential of function with polynomial growth, factorial or factorial-type function
coefficients grow or decay exponentially	$\lim \sup_{k \to \infty} \|a_k\|^{1/k} = \lambda$ ( $\lambda > 1$ if coefficients grow, $\lambda < 1$ if coefficients decay)		$1/\lambda$	To determine whether endpoints are included, we use the degree difference test or some variant (see below)	exponential function, exponential function times rational function times function of slower growth
coefficients grow or decay subexponentially	$\lim \sup_{k \to \infty} \|a_k\|^{1/k} = 1$		$1$	To determine whether endpoints are included, we use the degree difference test or some variant (see below)	exponential function, exponential function times rational function times function of slower growth

Explanation

The result follows almost immediately from the root test applied to the series $\sum_{k=0}^\infty a_kx^k$ . The key thing to remember is that the terms of this series are not $a_k$ , but $a_kx^k$ . We get:

$\lim \sup_{k \to \infty} |a_kx^k|^{1/k} = |x| \lim \sup_{k \to \infty} |a_k|^{1/k}$

The radius of convergence is the value $c$ which satisfies:

$c \lim \sup_{k \to \infty} |a_k|^{1/k} = 1$

This is because if $|x| > c$ , then $|x| \lim \sup_{k \to \infty} |a_k|^{1/k} > 1$ , so the series diverges by the root test. If $|x| < c$ , then $|x| \lim \sup_{k \to \infty} |a_k|^{1/k} < 1$ , so the series converges by the root test. The case $|x| = c$ (the endpoints) is inconclusive for the root test, and the rest of this page explains how to tackle the endpoints.

The table above simply considers the various cases for $\lim \sup |a_k|^{1/k}$ .

Hierarchy of functions

List of functions

The big-oh notation $O(f(x))$ means a function that is bounded between constant positive multiples of $f(x)$ for $x$ large enough.

The following are functions with superexponential growth (which becomes superexponential decay when they appear in the denominator). They are arranged in decreasing order of growth:

Double exponential functions, i.e., functions of the form $a^{b^x}$ where $a,b > 1$ .
Function of the form $\exp(O(x^r))$ , where $r > 1$ , i.e., functions growing exponentially in a superlinear power of $x$ . Within these, the larger the value of $r$ , the faster the growth.
Functions of the form $\exp(O(x \ln x))$ . The factorial function lives here.

The only functions with exponential growth are functions of the form $\exp(kx)$ , $k > 0$ . Their reciprocals, which are functions of the form $\exp(-kx), k > 0$ , give examples of functions of exponential decay.

Here are functions of subexponential growth (their reciprocals have subexponential decay):

Functions of the form $\exp(O(x^r))$ , $0 < r < 1$
Functions that grow like polynomials or power functions, i.e., $O(x^r), r > 0$ . The larger the value of $r$ , the faster the growth.
Functions that grow like polylogarithmic functions, i.e., $O((\ln x)^r), r > 0$ .

Interaction between functions

Interaction between superexponentials: The product of two functions with superexponential growth still has superexponential growth. For the quotient of two functions with superexponential growth, first figure out which one is higher in the hierarchy. If the numerator is higher in the hierarchy, then we get superexponential growth overall. If the denominator is higher in the hierarchy, then we get superexponential decay overall. if they are at the same place in the hierarchy, try doing algebraic simplification and rewrite and see what you get -- you may end up getting anything.
Interaction between superexponential and exponential/subexponential: The superexponential calls the shots.
Interaction between exponential and subexponential: We still get something exponential, but the subexponential part is relevant to determining endpoint behavior.

Remember that all these are functions in $k$

Before we go on to apply the ideas of this hierarchy of functions to the convergence problem of power series, we need to remind ourselves of the following: we are interested in the growth or decay of the power series coefficients $a_k$ , which are functions of the indexing variable $k$ , not $x$ . Thus, wherever, we see $x$ in the above hierarchy, we need to replace it by $k$ . For instance, $\sum 2^{3^k}x^k$ describes double exponential growth of coefficients.

Determination of interval of convergence

Superexponential growth and decay

(Note: the video begins with a general discussion of the rules, then proceeds to superexponential growth and decay)

We list below several examples:

Power series in $x$ with indexing variable $k$	What can we say about the coefficients as functions of $k$ ?	Conclusion about radius of convergence and interval of convergence
$\sum_{k=0}^\infty \frac{x^k}{2^{3^k}}$	superexponential decay (because a double exponential is in the denominator)	radius of convergence is $\infty$ , interval of convergence is $\R$ .
$\sum_{k=0}^\infty 2^{e^k} x^k$	superexponential growth (because a double exponential is in the numerator)	radius of convergence is 0, interval of convergence is $\{ 0 \}$
$\sum_{k=0}^\infty \frac{2^k(x - 7)^k}{3^{k^2}}$	superexponential decay (based on denominator)	radius of convergence is $\infty$ , interval of convergence is $\R$
$\sum_{k=0}^\infty \frac{(k^2 + 1)x^k}{2^{k^5}}$	superexponential decay (because numerator is subexponential, denominator is superexponential)	radius of convergence is $\infty$ , interval of convergence is $\R$
$\sum_{k=0}^\infty \frac{2^k(x + 5)^k}{k!}$	superexponential decay (numerator is exponential, denominator is superexponential)	radius of convergence is $\infty$ , interval of convergence is $\R$
$\sum_{k=0}^\infty \frac{2^{k^2}(x + 5)^k}{k!}$	superexponential growth (numerator dominates denominator)	radius of convergence is $0$ , interval of convergence is $\{ -5 \}$
$\sum_{k=0}^\infty \frac{2^{2^k}(x - 7)^k}{3^{k^2}}$	superexponential growth (because double exponential in numerator dominates exponential of square in denominator)	radius of convergence is 0, interval of convergence is $\{ 7 \}$

Subexponential growth and decay: degree difference test

Here the coefficients grow or decay subexponentially. Note that in this case, the radius of convergence is always 1. To determine the interval of convergence, we need to plug in the endpoints and use the various versions of the degree difference test. The degree difference test tells us that for a series with terms of the form:

$\frac{p(k)}{q(k)}x^k$

with $p,q$ polynomials or more generally sums of power functions, we have:

Case on degree difference	Alternative formulation of case using the fact that degrees are integers (in case they are polynomials)	Interval of convergence
$\! \deg q - \deg p > 1$	$\! \deg q - \deg p \ge 2$	$[-1,1]$ (absolute convergence on entire interval of convergence including at endpoints)
$\! 0 < \deg q - \deg p \le 1$	$\! \deg q - \deg p = 1$	$[-1,1)$ (conditional convergence at -1, absolute convergence in interior)
$\! \deg q - \deg p \le 0$	$\! \deg q - \deg p \le 0$	$(-1,1)$ (absolute convergence on entire interval of convergence)

In cases where the degree difference test is indeterminate, we may need to use the integral test.

With these rules, we can try determining the interval of convergence in a few cases:

Power series in $x$ with indexing variable $k$	Degree difference	Interval of convergence (radius of convergence is 1)
$\sum_{k=0}^\infty \frac{kx^k}{k^2 + 1}$	1	$[-1,1)$ (conditional convergence at -1)
$\sum_{k=3}^\infty \frac{(k^3 + 1)x^k}{k^2 - 4}$	-1	$(-1,1)$
$\sum_{k=0}^\infty \frac{k^2x^k}{k^4 + 1}$	2	$[-1,1]$ (absolute convergence at both endpoints)
$\sum_{k=0}^\infty \frac{k^{4/3}x^k}{k^3 + 1}$	5/3	$[-1,1]$ (absolute convergence at both endpoints)
$\sum_{k=0}^\infty \frac{k^{4/3}x^k}{k^2 + 1}$	2/3	$[-1,1)$ (conditional convergence at -1)
$\sum_{k=2}^\infty \frac{x^k}{k^2\ln(k^2 + 1)}$	$2^+$	$[-1,1]$ (absolute convergence at both endpoints)
$\sum_{k=2}^\infty \frac{x^k}{k[(\ln(k))^2 + 1]}$	$1^+$	the degree difference test is indeterminate, so we need to use the integral test for the endpoints. We obtain absolute convergence at both endpoints, so $[-1,1]$ (absolute convergence at both endpoints)
$\sum_{k=0}^\infty \frac{k^2(x + 4)^k}{k^3 + 1}$	1	$[-5,-3)$ (conditional convergence at -5) -- note that the series is centered at -4, so we need to shift our answer to that center.
$\sum_{k=1}^\infty \frac{x^k}{k\ln(k + 2)}$	$1^+$	the degree difference test is indeterminate, so we need to use the integral test for the endpoints. We obtain divergence at 1, so $[-1,1)$ (conditional convergence at -1)

Subexponential growth and decay: degree difference test with cautionary use where exponent is a multiple of the index of summation

We consider some examples for which we can use the degree difference test but need to do so more carefully.

Power series in $x$ with indexing variable $k$ (but note that the exponent is no longer precisely $k$ )	Degree difference	Interval of convergence (radius of convergence is 1)
$\sum_{k=0}^\infty \frac{kx^{2k}}{k^2 + 1}$	1	$(-1,1)$ : because we have only even powers appearing, it is now the case that at both $x = 1$ and $x = -1$ , we get a non-alternating series, which diverges. Hence, neither is in the interval of convergence.
$\sum_{k=0}^\infty \frac{kx^{2k+1}}{k^2 + 1}$	1	$(-1,1)$ : because we have only odd powers appearing, it is now the case that at both $x = 1$ and $x = -1$ , we get a non-alternating series, which diverges. Hence, neither is in the interval of convergence.
$\sum_{k=0}^\infty \frac{k(-1)^kx^{2k}}{k^2 + 1}$	1	$[-1,1]$ : because we now have alternating series at both endpoints.
$\sum_{k=0}^\infty \frac{kx^{3k}}{k^2 + 1}$	1	$[-1,1)$ : We have sign alternation for -1 but not for 1.

Subexponential growth and decay: examples where the exponent is a power of the index of summation

Power series in $x$ with indexing variable $k$ (but note that the exponent is no longer precisely $k$ )	Degree difference	Interval of convergence (radius of convergence is 1)
$\sum_{k=0}^\infty \frac{kx^{k^2}}{k^2 + 1}$	1	$[-1,1)$ (conditional convergence at $-1$ ): the key thing to note is that the sign alternation at -1 happens exactly as it would for $x^k$ instead of $x^{k^2}$ , so the analysis is the same as for $\sum_{k=0}^\infty \frac{kx^k}{k^2 + 1}$ .

(More need to be inserted to match the examples in the video).

Exponential growth or decay: degree difference test on coefficients

In this case, the coefficients grow or decay exponentially. The trick here is towrite the coefficient as a product of a pure exponential function (which determines the radius of convergence) and a subexponential function (which determines whether endpoints are included, via the degree difference test discussed above). Note that instead of $(-1,1), [-1,1), (-1,1],[-1,1]$ , we are now dealing with intervals specified using the radius of convergence.

Case on degree difference	Alternative formulation of case using the fact that degrees are integers (in case they are polynomials)	Interval of convergence for power series centered at 0 where radius of convergence is $c$ , which is the recirprocal of the exponential growth rate of coefficients
$\! \deg q - \deg p > 1$	$\! \deg q - \deg p \ge 2$	$[-c,c]$ (absolute convergence on entire interval of convergence including at endpoints)
$\! 0 < \deg q - \deg p \le 1$	$\! \deg q - \deg p = 1$	$[-c,c)$ (conditional convergence at $-c$ , absolute convergence in interior)
$\! \deg q - \deg p \le 0$	$\! \deg q - \deg p \le 0$	$(-c,c)$ (absolute convergence on entire interval of convergence)

We now consider some cases:

Power series in $x$ with indexing variable $k$	Number $\lambda$ such that coefficients are growing as $\lambda^k$	Radius of convergence is $1/\lambda$	Degree difference	Interval of convergence (radius of convergence is 1)
$\sum_{k=0}^\infty \frac{kx^k}{3^{2k}(k^2 + 1)}$	1/9	9	1	$[-9,9)$ (conditional convergence at -9)
$\sum_{k=3}^\infty \frac{(k^3 + 1)(2x)^k}{k^2 - 4}$	2	1/2	-1	$(-1/2,1/2)$
$\sum_{k=0}^\infty \frac{k^2(2x)^k}{5^{2k}(k^4 + 1)}$	2/25	25/2	2	$[-25/2,25/2]$ (absolute convergence at both endpoints)

Exponential growth or decay: cautionary use where the exponent is a multiple of the summation index

We consider cases where only some powers of $x$ appear. In these cases, we have to be careful determining the exponential growth rate and also the endpoint behavior.

For instance, for $\sum a^kx^{2k}$ , the exponential growth rate of coefficients is $\sqrt{|a|}$ and the radius of convergence is $1/\sqrt{|a|}$ .

Power series in $x$ with indexing variable $k$ (but note that the exponent is not precisely $k$ )	Exponential growth rate of coefficients	Radius of convergence	Degree difference	Interval of convergence (radius of convergence is 1)
$\sum_{k=0}^\infty \frac{k2^kx^{2k}}{k^2 + 1}$	$\sqrt{2}$	$1/\sqrt{2}$	1	$(-1/\sqrt{2},1/\sqrt{2})$ : because we have only even powers appearing, it is now the case that at both $x = 1/\sqrt{2}$ and $x = -1/\sqrt{2}$ , we get a non-alternating series, which diverges. Hence, neither is in the interval of convergence.
$\sum_{k=0}^\infty \frac{kx^{2k+1}}{3^k(k^2 + 1)}$	$1/\sqrt{3}$	$\sqrt{3}$	1	$(-\sqrt{3},\sqrt{3})$ : because we have only odd powers appearing, it is now the case that at both $x = \sqrt{3}$ and $x = -\sqrt{3}$ , we get a non-alternating series, which diverges. Hence, neither is in the interval of convergence.
$\sum_{k=0}^\infty \frac{k(-2)^kx^{2k}}{k^2 + 1}$	$\sqrt{2}$	$1/\sqrt{2}$	1	$[-1/\sqrt{2},1/\sqrt{2}]$ : because we now have alternating series at both endpoints.
$\sum_{k=0}^\infty \frac{5^kkx^{3k}}{k^2 + 1}$	$5^{1/3}$	$1/5^{1/3}$	1	$[-1/5^{1/3},1/5^{1/3})$ : We have sign alternation for the negative end but not for the positive end.

Tricky examples

Power series in $x$ with indexing variable $k$ (but note that the exponent is not precisely $k$ )	Analysis
$\sum_{k=0}^\infty \frac{2^kx^{k^2}}{k^2 + 1}$	The coefficients are actually growing subexponentially, not exponentially. The reason is that we need to compare the growth rate of the coefficients relative to $k^2$ , not $k$ . If we put $u = k^2$ , then the coefficient of $x^u$ is $\frac{2^{\sqrt{u}}}{u + 1}$ , which grows subexponentially in $u$ . Because this is a case of subexponential growth, the interval of convergence is $(-1,1)$ (note that because the coefficients are growing, convergence cannot occur at endpoints).

Power series in $x$ with indexing variable $k$ (but note that the exponent is not precisely $k$ )

Analysis

$\sum_{k=0}^\infty \frac{2^kx^{k^2}}{k^2 + 1}$

The coefficients are actually growing subexponentially, not exponentially. The reason is that we need to compare the growth rate of the coefficients relative to $k^2$ , not $k$ . If we put $u = k^2$ , then the coefficient of $x^u$ is $\frac{2^{\sqrt{u}}}{u + 1}$ , which grows subexponentially in $u$ . Because this is a case of subexponential growth, the interval of convergence is $(-1,1)$ (note that because the coefficients are growing, convergence cannot occur at endpoints).

(More examples to be added from video)