Positive-Series Thresholds, the Root Test, and the Radius of Convergence

Learning goals and dependencies

A convergence test is useful only after its hypotheses and its conclusion have been distinguished. This chapter proves integral and dyadic tests for nonnegative series, uses them to resolve logarithmic borderlines, and derives the root test and the power-series radius formula from the definition of limit superior.

We use completeness of the real numbers, monotone convergence for real sequences, the Cauchy criterion, elementary properties of logarithms, and proper and improper integration. A series beginning at a different finite index has the same convergence behavior: two such partial sums differ by one fixed finite sum. Its value, however, usually changes.

1. Positive terms: partial sums, comparisons, and tails

For $a_n\ge0$, write $A_N=\sum_{n=1}^N a_n$. These partial sums are nondecreasing. If bounded above, completeness gives $A_N\to\sup_N A_N$; if not bounded above, for every real $K$ some partial sum exceeds $K$, and all later partial sums do too. Thus a nonnegative series either has a finite sum or diverges to $+\infty$.

If $0\le a_n\le b_n$ for all $n\ge n_0$ and $\sum b_n$ converges, the partial sums of the tail of $a_n$ are bounded by the finite sum of the corresponding $b_n$ tail. They therefore converge. If instead $\sum a_n=+\infty$, the larger tail partial sums of $b_n$ are unbounded. Notice the direction: divergence of the larger series does not decide the smaller one.

Limit comparison. Suppose $a_n,b_n>0$ eventually and $a_n/b_n\to\ell$ with $0<\ell<\infty$. Choose an index beyond which $\ell/2<a_n/b_n<3\ell/2$. Then

$$\frac\ell2 b_n\le a_n\le\frac{3\ell}2b_n.$$

Multiplication by a positive constant does not affect boundedness of nonnegative partial sums. The two comparison directions prove that the series converge or diverge together. The assumptions $\ell>0$ and $\ell<\infty$ cannot simply be deleted: $a_n/b_n\to0$ gives only the upper comparison needed when $\sum b_n$ converges, not an equivalence.

For a convergent series, the remainder after $m$ terms is

$$R_m=\sum_{n=m+1}^{\infty}a_n.$$

Proving convergence and giving a numerical bound for $R_m$ are different achievements. The next test provides both.

2. The integral test, including its inequalities

Theorem 1. Let $f:[N,\infty)\to[0,\infty)$ be nonincreasing, where $N$ is a positive integer. Then $\sum_{n=N}^{\infty}f(n)$ converges if and only if $\int_N^{\infty}f(x)\,dx$ is finite. When they converge, for every integer $m\ge N$,

$$\int_{m+1}^{\infty}f(x)\,dx\le\sum_{n=m+1}^{\infty}f(n)\le\int_m^{\infty}f(x)\,dx.$$

A bounded monotone function is Riemann integrable on every compact subinterval, so no continuity assumption has been hidden here.

Proof. On $[n,n+1]$, monotonicity gives $f(n+1)\le f(x)\le f(n)$. Integrating this pointwise inequality gives

$$f(n+1)\le\int_n^{n+1}f(x)\,dx\le f(n).$$

Adding the right inequality from $n=N$ to $M$ gives $\int_N^{M+1}f\le\sum_{n=N}^{M}f(n)$. Adding the left inequality from $n=N$ to $M-1$ gives $\sum_{n=N+1}^{M}f(n)\le\int_N^M f$. Consequently

$$\int_N^{M+1}f\le\sum_{n=N}^{M}f(n)\le f(N)+\int_N^M f.$$

If the improper integral is finite, the right side bounds all partial sums. If the series converges, the left inequality bounds $\int_N^{M+1}f$. For an arbitrary real upper endpoint $T$, choose an integer $M$ with $T\le M+1$; nonnegativity then bounds $\int_N^T f$ as well. The nondecreasing integral therefore has a finite limit. This proves the equivalence.

For the remainder, apply the right rectangle bound on $[n,n+1]$ for $n=m+1,m+2,\ldots$ to get the lower integral bound. Apply the left rectangle bound on $[n-1,n]$ for those same integers to get the upper integral bound. First add finitely many inequalities and then pass to their limits; all quantities converge and preserve the inequality. $\square$

Worked example 1: the complete $p$-series threshold. If $p\le0$, $n^{-p}\ge1$, so the terms do not tend to zero. If $p>0$, $x^{-p}$ is positive and decreasing. For $p\ne1$,

$$\int_1^T x^{-p}\,dx=\frac{T^{1-p}-1}{1-p},$$

and for $p=1$ the integral is $\log T$. These expressions have a finite limit precisely when $p>1$. Therefore $\sum n^{-p}$ converges exactly for $p>1$. In that case the remainder satisfies

$$\frac{(m+1)^{1-p}}{p-1}\le R_m\le\frac{m^{1-p}}{p-1}.$$

For $p=3$, this is $1/[2(m+1)^2]\le R_m\le1/(2m^2)$. For example $m=100$ certifies $R_{100}\le1/20000$; it does not certify that the exact remainder equals that upper bound.

Worked example 2: a logarithm changes the borderline. Consider

$$\sum_{n=3}^{\infty}\frac1{n(\log n)^q},\qquad q\in\mathbb R.$$

For $f(x)=1/[x(\log x)^q]$ with $x>1$,

$$\frac{f'(x)}{f(x)}=-\frac1x-\frac{q}{x\log x}=-\frac{1+q/\log x}{x}.$$

For each fixed real $q$, the numerator $1+q/\log x$ is positive for all sufficiently large $x$. Thus the integral test is applicable to a tail even if $f$ is not decreasing at the initial index. Substituting $u=\log x$ gives

$$\int\frac{dx}{x(\log x)^q}=\int u^{-q}\,du.$$

The integral from any fixed positive lower $u$ to infinity is finite exactly when $q>1$. Hence this logarithmic series converges exactly when $q>1$. In particular $\sum1/(n\log n)$ diverges, although its terms are smaller than the harmonic terms. That smaller-term observation alone could never have established convergence.

For $q>1$ and sufficiently large $m$, its remainder is at most

$$\int_m^{\infty}\frac{dx}{x(\log x)^q}=\frac{(\log m)^{1-q}}{q-1}.$$

At $q=2$ this decreases only as $1/\log m$, a warning that a convergent series can converge very slowly.

3. Dyadic condensation: the monotonicity is doing work

Theorem 2. If $a_n\ge0$ and $a_n$ is nonincreasing, then

$$\sum_{n=1}^{\infty}a_n\quad\hbox{converges if and only if}\quad\sum_{k=0}^{\infty}2^k a_{2^k}\quad\hbox{converges}.$$

Proof. The block $B_k=\sum_{n=2^k}^{2^{k+1}-1}a_n$ contains exactly $2^k$ terms. Every term is at most $a_{2^k}$ and at least $a_{2^{k+1}}$, so

$$\frac12\bigl(2^{k+1}a_{2^{k+1}}\bigr)\le B_k\le2^k a_{2^k}.$$

If the condensed series converges, the upper bound makes the partial sums of the block series bounded. If the block series converges, the lower bound makes the condensed series from $k=1$ onward bounded. Adding its one missing term does not change convergence. Finally, the block endpoints are a subsequence of the original nonnegative partial sums. Bounded block-endpoint sums bound every intermediate partial sum, because it lies between neighboring endpoints. Hence the block series converges precisely when the original series does. $\square$

If monotonicity holds only from some index onward, replace the finitely many earlier $a_n$ by the first value of the monotone tail. This produces a globally nonincreasing nonnegative sequence while changing only finitely many terms of both series. The same conclusion follows.

Worked example 3: change the logarithm's base, not the threshold. For $q>0$, $a_n=1/[n(\log_2 n)^q]$ decreases for $n\ge2$. At a dyadic index,

$$2^k a_{2^k}=\frac1{k^q}\qquad(k\ge1).$$

Condensation and the $p$-series theorem give convergence exactly when $q>1$. If $q\le0$, then $(\log_2 n)^q\le1$ for $n\ge2$, so $a_n\ge1/n$ and the series diverges. Thus the same threshold holds for every real $q$. This agrees with the integral calculation, since $\log_2 n=(\log n)/(\log2)$ changes each term by one positive constant factor.

4. The limit superior is a tail bound, not a guessed limit

For a sequence $u_n\ge0$, define

$$s_N=\sup\{u_n:n\ge N\},\qquad L=\inf_N s_N=\limsup_{n\to\infty}u_n,$$

allowing the value $+\infty$. The tail suprema are nonincreasing, so their infimum is their extended limit. Two consequences are needed.

• If $L<q$, then $u_n<q$ eventually. Indeed some tail supremum is below $q$; otherwise their infimum would be at least $q$.
• If $q<L$, then $u_n>q$ for infinitely many indices. Otherwise a tail would have every term at most $q$, forcing its supremum and hence $L$ to be at most $q$.

These statements remain useful when the original sequence has no limit. A finite number of unusually large terms affects some early tail suprema, but not $L$.

Theorem 3: root test. For real terms $a_n$, let $L=\limsup |a_n|^{1/n}$, using $0^{1/n}=0$.

1. If $L<1$, $\sum a_n$ converges absolutely.
2. If $L>1$, including $L=+\infty$, $a_n$ does not tend to zero, so $\sum a_n$ diverges.
3. If $L=1$, neither convergence nor divergence is determined by this test.

Proof. In the first case choose $q$ with $L<q<1$. Eventually $|a_n|^{1/n}<q$, giving $|a_n|<q^n$. Comparison with a convergent geometric series proves absolute convergence.

In the second case choose a finite $q$ with $1<q<L$; this choice is possible also when $L=+\infty$. Infinitely many indices satisfy $|a_n|^{1/n}>q$, so at those indices $|a_n|>q^n\ge1$. Consequently $a_n$ cannot tend to zero.

For the third case, $\sum1/n^2$ converges, $\sum1/n$ diverges, and the alternating harmonic series converges conditionally. In all three cases $|a_n|^{1/n}\to1$. To verify the root limit without assuming it, first observe that for every $\varepsilon>0$,

$$(1+\varepsilon)^n\ge\binom n2\varepsilon^2\ge n$$

when $n\ge1+2/\varepsilon^2$. Thus $1\le n^{1/n}\le1+\varepsilon$ eventually, and $n^{1/n}\to1$. Taking its reciprocal or its reciprocal square gives the required limits. $\square$

A root limit equal to one is not evidence of divergence. It says that this particular geometric comparison has reached its decision boundary.

5. Deriving the radius formula

Consider the real power series

$$\sum_{n=0}^{\infty}c_n(x-a)^n,$$

and define $L=\limsup_{n\to\infty}|c_n|^{1/n}$. The constant coefficient $c_0$ does not enter this definition. For $t=|x-a|>0$,

$$\limsup |c_n(x-a)^n|^{1/n}=tL.$$

For finite $L$, this follows because multiplication of every tail supremum by the positive number $t$ commutes with taking the infimum. If $L=+\infty$, all those tail suprema are infinite, so multiplying by positive $t$ still gives infinity.

The root test now gives the Cauchy–Hadamard formula

$$R=\frac1L,\qquad\frac10=+\infty,\qquad\frac1{+\infty}=0.$$

If $0<L<\infty$, the series converges absolutely when $t<R$ and diverges when $t>R$. If $L=0$, it converges absolutely for every finite $t$, so $R=+\infty$. If $L=+\infty$, it diverges at every $t>0$, so $R=0$. At the center $x=a$, every term with $n\ge1$ is zero, and the sum is $c_0$ in all cases. When $0<R<\infty$, the endpoints have $tL=1$; the root test deliberately leaves them undecided.

Worked example 4: zero coefficients defeat naive ratios, not the root test. Let $c_0=0$, and for $k\ge1$ set

$$c_{2k}=\frac{3^{2k}}{(2k)^2},\qquad c_{2k-1}=0.$$

Along even indices, $|c_{2k}|^{1/(2k)}=3/(2k)^{1/k}\to3$; along odd indices the root is zero. Thus the limit does not exist, but the limit superior is $3$, and $R=1/3$. Ratios of successive coefficients are not defined when the denominator is zero. At either endpoint $x=\pm1/3$, only even powers remain, and the absolute series is $\sum_{k\ge1}1/(2k)^2$, which converges. The interval of convergence is therefore $[-1/3,1/3]$.

Worked example 5: one endpoint converges and the other does not. For

$$\sum_{n=2}^{\infty}\frac{2^n x^n}{n\log n},$$

we have $1\le n\log n\le n^2$ eventually. Its $n$th root tends to one by squeezing between $1$ and $n^{2/n}$, so the coefficient-root limit is $2$ and $R=1/2$. At $x=1/2$ the logarithmic series of Worked example 2 diverges. At $x=-1/2$ the terms are $(-1)^n/(n\log n)$; the positive magnitudes decrease to zero, so the alternating test gives convergence. Its absolute series diverges. Thus the convergence interval is $[-1/2,1/2)$, with conditional convergence at the left endpoint.

6. Graded exercises

Foundation

P01. Give explicit bounds for $\sum_{n=m+1}^{\infty}n^{-4}$. Find an integer $m$ that certifies the remainder is at most $10^{-6}$.

P02. Decide convergence of $\sum_{n=3}^{\infty}1/[n(\log n)^{3/2}]$ and $\sum_{n=3}^{\infty}1/[n(\log n)^{1/2}]$. State the substitution and the tail monotonicity needed for the integral test.

P03. If $a_n,b_n>0$ and $a_n/b_n\to0$, does divergence of $\sum b_n$ imply divergence of $\sum a_n$? Supply one counterexample in which $\sum a_n$ converges and one in which it diverges.

Core

P04. Prove that condensation cannot be used without monotonicity by taking $a_n=1/n$ at powers of two and $a_n=0$ at all other indices. Compute both series' behavior, and verify that $a_n\to0$.

P05. Let $a_n=4^{-n}$ for even $n$ and $a_n=9^{-n}$ for odd $n$. Compute the limit superior and limit inferior of $|a_n|^{1/n}$. Decide convergence using the root test and explain why a single root limit would be an incorrect description.

P06. Prove directly that $\sum_{n=1}^{\infty}n^2/5^n$ converges by the root test. Your solution must justify the root limit, not merely state that exponentials grow faster than polynomials.

P07. Let $c_0=0$, $c_n=n^2 2^n$ if $3$ divides $n$, and $c_n=0$ otherwise. Find the radius and the full real convergence interval of $\sum c_nx^n$.

Synthesis and proof audit

P08. Let $c_n=5^n$ if $n=2^k$ for an integer $k\ge1$, and $c_n=0$ otherwise, including $n=0,1$. Find the radius and check both endpoints. Explain why having arbitrarily long strings of zero coefficients does not imply an infinite radius.

P09. Suppose two power series about the same center differ in only finitely many coefficients. Prove that they have the same radius and the same endpoint convergence behavior, although their sums need not agree.

P10. Determine all real $q$ for which

$$\sum_{n=16}^{\infty}\frac1{n\log n\,(\log\log n)^q}$$

converges. Verify the sign and eventual monotonicity conditions before applying an integral substitution.

7. Full solutions

S01. Apply the remainder inequalities to $f(x)=x^{-4}$. Integration gives

$$\frac1{3(m+1)^3}\le\sum_{n=m+1}^{\infty}\frac1{n^4}\le\frac1{3m^3}.$$

We need $3m^3\ge10^6$. The choice $m=70$ gives $3m^3=1{,}029{,}000$, so the upper bound is $1/1{,}029{,}000<10^{-6}$. This is a sufficient certified choice, without a claim that it is the smallest index for the actual remainder.

S02. For each positive exponent $q$, $f'(x)/f(x)=-(1+q/\log x)/x<0$ on $x>1$. Both integrands are positive for $x\ge3$. Set $u=\log x$, so $du=dx/x$. For $q=3/2$, the tail integral is $\int u^{-3/2}du$, which is finite at infinity; its value from $\log3$ onward is $2/\sqrt{\log3}$. The series converges. For $q=1/2$, an antiderivative is $2\sqrt u$, which grows without bound, so the integral and the series diverge.

S03. No. Set $b_n=1/n$. With $a_n=1/n^2$, the ratio is $1/n\to0$ but the smaller series converges. With $a_n=1/[n\log(n+1)]$, the ratio is $1/\log(n+1)\to0$ and the smaller series still diverges. To justify the latter, for $n\ge2$ we have $n+1\le n^2$, so $\log(n+1)\le2\log n$. Hence $a_n\ge1/(2n\log n)$, and the divergent logarithmic series provides the lower comparison. The limit zero provides no fixed positive lower multiple of $b_n$.

S04. At $n=2^k$, $a_n=2^{-k}$, so the original series equals $\sum_{k=0}^{\infty}2^{-k}=2$. At other indices the terms vanish, and in every case $0\le a_n\le1/n$, so $a_n\to0$. But $2^ka_{2^k}=1$ for every $k$, making the condensed series divergent. The sequence is not nonincreasing: immediately before a sufficiently large power of two it jumps from zero to a positive value. This violates the exact hypothesis that made the block inequalities possible.

S05. The root is $1/4$ at every even index and $1/9$ at every odd index. Every tail contains both kinds of index, so its supremum is $1/4$ and its infimum is $1/9$. Therefore $\limsup=1/4$ and $\liminf=1/9$. The root test gives absolute convergence because $1/4<1$. The unequal subsequential limits show that the root sequence itself has no limit.

S06. We have

$$\left(\frac{n^2}{5^n}\right)^{1/n}=\frac{n^{2/n}}5.$$

The binomial argument in Theorem 3 proves $n^{1/n}\to1$, so its square tends to one as well. The displayed roots tend to $1/5<1$, and the root test proves convergence. This proof includes the quantitative reason the polynomial factor disappears under $n$th roots.

S07. On multiples of three, the coefficient root is $2n^{2/n}\to2$; on all other indices it is zero. To see that the limit superior is exactly $2$, note that all sufficiently large nonzero roots are below $2+\varepsilon$, while infinitely many of them are above $2-\varepsilon$ for every $\varepsilon>0$. Hence $R=1/2$. At $x=1/2$, the terms on multiples of three are $n^2$; they do not tend to zero. At $x=-1/2$, the corresponding terms are $(-1)^n n^2$, whose magnitudes again do not tend to zero. Both endpoints diverge. The convergence interval is $(-1/2,1/2)$.

S08. At each selected index the coefficient root is $5$; elsewhere it is zero. Every tail contains a selected index, so the tail supremum is $5$ and the radius is $1/5$. At $x=1/5$, the selected terms are all one, preventing the term limit from being zero. At $x=-1/5$, the selected exponents $2^k$ are even, so these terms are again one. The interval is $(-1/5,1/5)$. Long zero strings do not change the fact that every tail has root supremum $5$; the radius depends on tail suprema, not on the proportion of nonzero coefficients.

S09. Let the coefficients agree for all $n>K$. The two coefficient-root sequences therefore have identical tails once $N>K$, giving identical limits superior and identical radii by the proved formula. At any fixed real input, the two series differ by the finite polynomial $\sum_{n=0}^K(c_n-d_n)(x-a)^n$. Their partial-sum difference is that fixed finite value for all sufficiently large truncation indices. One series converges at the input if and only if the other does. This includes any finite endpoints. The difference polynomial need not be zero, so equality of sums is not implied.

S10. For $x\ge16>e$, $\log x>0$ and $\log\log x>0$, so the expression is positive for every real $q$. Its logarithmic derivative is

$$-\frac1x-\frac1{x\log x}-\frac{q}{x\log x\log\log x} =-\frac1x\left(1+\frac1{\log x}+\frac{q}{\log x\log\log x}\right).$$

The bracket tends to one and is therefore positive eventually, proving eventual decrease. Take $u=\log\log x$, so $du=dx/(x\log x)$. The improper tail integral becomes $\int u^{-q}du$ from a positive lower bound to infinity. It is finite if $q>1$, equals a logarithm diverging to infinity if $q=1$, and grows as a positive power of the upper endpoint if $q<1$. The integral test on the decreasing tail proves that the original series converges exactly for $q>1$.

Summary

Comparison controls nonnegative partial sums. Integral and condensation tests require a monotone structure, and their proofs explain why. The root test uses an eventual upper bound or an infinite sequence of lower bounds furnished by limit superior. The Cauchy–Hadamard formula is then a consequence, with zero, infinite, and endpoint cases handled separately. A radius calculation alone determines neither endpoint sums nor whether infinite operations may be interchanged.