From Root Bounds to a Proof
The Root Test classifies a series by the limit superior of the nth roots of its term magnitudes. In the previous tutorial, we saw why a limit superior below one gives an eventual geometric bound. Here we complete the argument carefully: we connect that bound to absolute convergence, and show why a limit superior above one forces terms that fail to tend to zero.
The proof uses two different consequences of the limit superior. Below one, all sufficiently late roots are bounded by a fixed number less than one. Above one, roots exceed a fixed number greater than one at infinitely many indices. The second consequence is worth stating explicitly, especially when the root sequence does not converge.
Proof. Suppose instead that \(x_n>c\) for only finitely many indices. Then there is an index \(N\) such that \(x_n\leq c\) for every \(n\geq N\). Consequently, every tail supremum satisfies $$ \sup_{n\geq M}x_n\leq c\qquad(M\geq N). $$ Taking the infimum of the tail suprema gives \(\limsup_{n\to\infty}x_n\leq c\), contradicting the assumption \(c<\limsup_{n\to\infty}x_n\). Thus \(x_n>c\) infinitely often. \(\square\)
Proof of the Root Test
Write \(x_n=|a_n|^{1/n}\), with \(0^{1/n}=0\). First suppose \(\rho<1\). Choose a real number \(q\) such that $$ \rho<q<1. $$ By the proposition “A Root Bound Gives a Geometric Bound” from the previous tutorial, there is an index \(N\) such that $$ |a_n|\leq q^n\qquad(n\geq N). $$ The geometric series \(\sum_{n=N}^{\infty}q^n\) converges because \(0<q<1\). The Comparison Test for Positive-Term Series therefore shows that \(\sum_{n=N}^{\infty}|a_n|\) converges. Adding the finitely many terms with \(n<N\) does not affect convergence, so \(\sum_{n=1}^{\infty}|a_n|\) converges. This is precisely absolute convergence of \(\sum_{n=1}^{\infty}a_n\).
Now suppose \(1<\rho\leq+\infty\). Choose a finite number \(c>1\) with \(c<\rho\); this is possible when \(\rho\) is finite and greater than one, and when \(\rho=+\infty\) we may, for example, take \(c=2\). Apply the lemma to \(x_n=|a_n|^{1/n}\). It follows that $$ |a_n|^{1/n}>c $$ for infinitely many indices \(n\). Since \(c>1\), at each such index $$ |a_n|>c^n\geq c>1. $$ Thus the terms \(a_n\) do not tend to zero. The Necessary Condition for Series Convergence says that the terms of a convergent series must tend to zero. Therefore \(\sum_{n=1}^{\infty}a_n\) diverges.
These arguments also show why the cases lie on opposite sides of one. Below one, a fixed geometric sequence controls the tail. Above one, infinitely many terms have magnitude bounded below by a number greater than one. At exactly one, neither conclusion follows from the root information alone; examples below will show both possible outcomes.
A Quantitative Bound for the Convergent Tail
The convergence proof gives more than absolute convergence. Once a geometric bound is available, it also bounds the size of the tail starting at any sufficiently late index. This estimate can be useful when the Root Test is applied to a series whose remainder needs to be controlled.
Proof. By “A Root Bound Gives a Geometric Bound,” choose \(N\) so that \(|a_n|\leq q^n\) for every \(n\geq N\). If \(M\geq N\), this inequality holds for every \(n\geq M\). Applying it first to finite sums and then taking the limit gives $$ \sum_{n=M}^{\infty}|a_n| \leq\sum_{n=M}^{\infty}q^n =\frac{q^M}{1-q}. $$ The last equality is the geometric series formula. This proves the bound. \(\square\)
The value of \(q\) need not equal \(\rho\), and usually should not be treated as though it does. The Root Test guarantees a suitable fixed \(q\) strictly between \(\rho\) and one; that is what makes the geometric tail estimate valid. A smaller available value of \(q\) gives a sharper bound, but it must still satisfy the eventual root bound.
Worked Applications
Worked Example: A Root Bound and an Explicit Remainder Estimate
For \(n\geq1\), define $$ a_n=\left(\frac25\right)^n\left(1+\frac1n\right)^n. $$ Taking the nth root gives the exact identity $$ |a_n|^{1/n}=\frac25\left(1+\frac1n\right). $$ As \(n\to\infty\), this tends to \(2/5\), so \(\rho=2/5<1\). The Root Test proves absolute convergence.
We can also obtain a concrete tail estimate. For every \(n\geq4\), \(1+1/n\leq5/4\), and therefore $$ |a_n|^{1/n} =\frac25\left(1+\frac1n\right) \leq\frac25\cdot\frac54 =\frac12. $$ Raising both sides to the power \(n\) gives \(|a_n|\leq(1/2)^n\) for \(n\geq4\). Hence, for every \(M\geq4\), $$ \sum_{n=M}^{\infty}|a_n| \leq\sum_{n=M}^{\infty}\left(\frac12\right)^n =2^{1-M}. $$ This provides both a convergence proof and an explicit bound on the omitted tail.
Worked Example: Large Terms at Square Indices
Define a sequence by $$ a_n= \begin{cases} 3^n,&\text{if }n\text{ is a perfect square},\\ 0,&\text{otherwise}. \end{cases} $$ For \(n=k^2\), the nth root is $$ |a_{k^2}|^{1/k^2}=(3^{k^2})^{1/k^2}=3. $$ At every nonsquare index, the nth root is zero. Since there are infinitely many square indices, the root sequence equals \(3\) infinitely often and is always at most \(3\). Thus $$ \limsup_{n\to\infty}|a_n|^{1/n}=3>1. $$ The divergence case of the Root Test applies. Directly, the terms at square indices satisfy \(a_{k^2}=3^{k^2}\), so the terms do not tend to zero. This example shows that zeros at many indices do not prevent the divergence case from applying: the limit superior records what happens along the indices where the roots are large.
Worked Example: A Root Limit of One with a Convergent Series
Consider the positive terms $$ a_n=\frac{1}{n(n+1)}. $$ Their nth roots satisfy $$ |a_n|^{1/n} =\exp\left(-\frac{\log n+\log(n+1)}{n}\right) \longrightarrow1, $$ because \(\log n/n\to0\) and \(\log(n+1)/n\to0\). Thus the Root Test reaches its boundary value.
Nevertheless, the series converges. The identity $$ \frac{1}{n(n+1)}=\frac1n-\frac{1}{n+1} $$ gives, for every positive integer \(N\), $$ \sum_{n=1}^{N}\frac{1}{n(n+1)} =\sum_{n=1}^{N}\left(\frac1n-\frac{1}{n+1}\right) =1-\frac{1}{N+1}. $$ As \(N\to\infty\), these partial sums tend to \(1\). Thus the series converges even though the root limit is one.
Worked Example: A Root Limit of One with a Divergent Series
Now take \(b_n=1/\sqrt n\). Its nth root is $$ |b_n|^{1/n} =\exp\left(-\frac{\log n}{2n}\right) \longrightarrow1. $$ The Root Test again gives the boundary value. But \(\sum_{n=1}^{\infty}1/\sqrt n\) is the p-series with \(p=1/2\), so it diverges by the p-Series Convergence Criterion. The two boundary examples have the same root limit and different convergence behavior. Therefore a value of one cannot be turned into either a convergence or a divergence conclusion without additional information.
What the Proof Does—and Does Not—Say
The key distinction is whether the root limit superior is separated from one. When \(\rho<1\), the gap allows a choice of \(q<1\) that bounds all sufficiently late roots, and the resulting geometric comparison controls the whole tail. When \(\rho>1\), the limit superior guarantees infinitely many roots above some \(c>1\), which contradicts the necessary condition that terms tend to zero.
A common pitfall is to treat the boundary value as a weak version of one of the other cases. It is not. The convergent telescoping series and the divergent p-series above both have root limit one. The test is inconclusive there, even when the ordinary root limit exists. A different argument or a different test is needed.
Check Your Understanding
Use the proof and examples to answer the following questions.
- Why does \(\limsup x_n>c\) imply that \(x_n>c\) for infinitely many indices?
- In the convergence proof, where is the condition \(q<1\) used?
- Why do the square-index terms in the sparse sequence force divergence?
- What tail bound follows when \(|a_n|\leq q^n\) for all \(n\geq N\), with \(0<q<1\)?
- Give one reason that a root limit of one does not determine whether a series converges.