Finding Parameter Thresholds with the Root Test

Why Parameter Thresholds Need Boundary Checks

A series may contain a parameter in a base raised to the index, such as \(b(u)^n\). The Root Test is well suited to such terms: taking the \(n\)th root removes the exponent \(n\) and leaves a condition on the parameter. Solving that condition identifies ranges where the test guarantees convergence or divergence.

The threshold itself needs separate attention. When the Root Test limit equals \(1\), the test gives no conclusion. A series at that parameter value may converge or diverge, depending on the remaining factors in its terms. The aim is therefore not just to find a threshold, but to distinguish the strict cases from the boundary cases.

The Root Test and Parameter Thresholds

Theorem: For a series \(\sum_{n=1}^{\infty} a_n\), let \(L=\limsup_{n\to\infty}\sqrt[n]{|a_n|}\). If \(L<1\), the series converges absolutely. If \(L>1\), the series diverges. If \(L=1\), the Root Test is inconclusive.

For a parameterized series, compute \(L\) as a function of the parameter. The inequality \(L<1\) identifies the parameter values for which the test proves convergence; \(L>1\) identifies values for which it proves divergence. The equation \(L=1\) marks candidate boundary values, not a conclusion about convergence.

The typical simplification comes from terms of the form \(c_n b(u)^n\), where \(u\) is the parameter and \(c_n\) is a factor whose \(n\)th root tends to \(1\). In that case,

\sqrt[n]{|c_n b(u)^n|} = \sqrt[n]{|c_n|}\,|b(u)|.

If \(\sqrt[n]{|c_n|}\to 1\), the Root Test limit is \(|b(u)|\). For example, when \(c_n=1/n^p\) with \(p>0\),

\sqrt[n]{\frac{1}{n^p}} = \exp\left(-\frac{p\ln n}{n}\right) \longrightarrow 1,

because \(\ln n/n\to 0\). Likewise, for \(c_n=n^r\), where \(r\) is fixed, \(\sqrt[n]{n^r}=\exp(r\ln n/n)\to 1\). Thus polynomial factors can affect what happens at a boundary even though they do not move the Root Test threshold.

Why the Root Test Gives These Strict Conclusions

The two strict cases follow from how a sequence behaves when its \(n\)th roots are eventually bounded on one side of \(1\). Here is the justification. It also explains why equality cannot be folded into either case.

When \(L<1\).
Choose a real number \(r\) with \(L<r<1\). By the definition of limit superior, for all sufficiently large \(n\), \(\sqrt[n]{|a_n|}<r\), so \(|a_n|<r^n\). The geometric series \(\sum r^n\) converges, and comparison shows that \(\sum |a_n|\) converges.

When \(L>1\).
Choose \(r\) with \(1<r<L\). Infinitely many \(n\) have \(\sqrt[n]{|a_n|}>r\), so for those indices \(|a_n|>r^n>1\). In particular, \(a_n\) does not tend to zero. Since convergence of a series requires its terms to tend to zero, the series diverges.

When \(L=1\).
There is no number \(r<1\) with \(L<r\), and no number \(r>1\) with \(r<L\). The comparison arguments above do not apply. The terms at the boundary must be examined by another argument.

Worked Example: A Convergent Boundary with a Reciprocal-Square Factor

Worked Example: A Reciprocal-Square Factor

For \(u\ge 0\), determine the convergence range of

\sum_{n=1}^{\infty}\frac{1}{n^2}\left(\frac{2u}{u+2}\right)^n.

For each fixed \(u\ge 0\), the base \(2u/(u+2)\) is nonnegative, and the \(n\)th root of the absolute value of the \(n\)th term is

\sqrt[n]{\frac{1}{n^2}\left|\frac{2u}{u+2}\right|^n} = \frac{1}{n^{2/n}}\frac{2u}{u+2}.

Since \(n^{2/n}\to 1\), the Root Test limit is \(L=2u/(u+2)\). The denominator is positive on the stated domain, so the strict convergence condition is

\frac{2u}{u+2}<1 \quad\Longleftrightarrow\quad 2u<u+2 \quad\Longleftrightarrow\quad u<2.

Thus the Root Test proves convergence for \(0\le u<2\). For \(u>2\), \(L>1\), so it proves divergence.

At \(u=2\), the base is \(2(2)/(2+2)=1\). Substitution gives the boundary series

\sum_{n=1}^{\infty}\frac{1}{n^2}(1)^n = \sum_{n=1}^{\infty}\frac{1}{n^2},

which converges by the \(p\)-series test with \(p=2\). The full conclusion is convergence for \(0\le u\le 2\) and divergence for \(u>2\).

As a numerical check, at \(u=1\) and \(n=2\), the base is \(2/3\), while the prefactor is \(1/2^2=1/4\). The term is therefore

\frac{1}{4}\left(\frac{2}{3}\right)^2 = \frac{1}{4}\cdot\frac{4}{9} = \frac{1}{9}.

This agrees with direct substitution into the original term.

Worked Example: A Boundary That Diverges

Worked Example: A Reciprocal Factor

For \(x\ge 0\), determine the convergence range of

\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{x}{2}\right)^n.

The \(n\)th root of the absolute value of the term is

\sqrt[n]{\frac{1}{n}\left(\frac{x}{2}\right)^n} = \frac{1}{n^{1/n}}\frac{x}{2} \longrightarrow \frac{x}{2},

because \(n^{1/n}\to 1\). The Root Test proves convergence when \(x/2<1\), or \(0\le x<2\), and proves divergence when \(x/2>1\), or \(x>2\).

At the threshold \(x=2\), the base is \(1\), and substitution gives

\sum_{n=1}^{\infty}\frac{1}{n}(1)^n = \sum_{n=1}^{\infty}\frac{1}{n}.

This is the harmonic series, which diverges. Therefore this series converges for \(0\le x<2\) and diverges for \(x\ge 2\). The same Root Test threshold as in the previous example has a different boundary outcome because the remaining factor is \(1/n\), not \(1/n^2\).

For a direct term check at \(x=2\), the \(n\)th term is \(1/n\), which tends to zero. That fact alone does not prove convergence; the harmonic series still diverges. This is one reason a boundary check must use an appropriate series test, rather than only checking whether the terms tend to zero.

Worked Example: An Absolute-Value Inequality with Two Boundary Values

Worked Example: A Linear Parameter in the Base

For real \(t\), determine the convergence range of

\sum_{n=1}^{\infty}n\left(\frac{3t-1}{5}\right)^n.

The \(n\)th root of the absolute value of the term is

\sqrt[n]{n\left|\frac{3t-1}{5}\right|^n} = n^{1/n}\left|\frac{3t-1}{5}\right| \longrightarrow \left|\frac{3t-1}{5}\right|.

The strict convergence condition is

\left|\frac{3t-1}{5}\right|<1 \quad\Longleftrightarrow\quad |3t-1|<5 \quad\Longleftrightarrow\quad -5<3t-1<5.

Adding \(1\) throughout gives \(-4<3t<6\), and dividing by \(3\) gives \(-4/3<t<2\). The Root Test therefore proves convergence inside this interval. Outside its closure, the root limit exceeds \(1\), so the series diverges.

At \(t=2\), the base is \((6-1)/5=1\), so the terms are \(n\), which do not tend to zero. At \(t=-4/3\), the base is \((-4-1)/5=-1\), so the terms are \(n(-1)^n\), whose absolute values are \(n\) and also do not tend to zero. The series diverges at both boundaries by the term test. Hence it converges precisely for \(-4/3<t<2\).

For instance, at \(t=1\) and \(n=2\), the base is \(2/5\), and the term is \(2(2/5)^2=2(4/25)=8/25\). Substituting this value into the displayed series term gives the same result.

A Reliable Procedure and a Common Pitfall

For a parameterized series with an index-dependent power, keep the threshold calculation and the boundary analysis as separate tasks:

Take the \(n\)th root of the absolute value of the general term.
Find its limit or limit superior as \(n\to\infty\), retaining the parameter.
Solve the strict inequality \(L<1\) for convergence and \(L>1\) for divergence.
Substitute each parameter value where \(L=1\) into the original series.
Use a suitable test on each resulting boundary series; do not infer its behavior from the Root Test.

A common mistake is to write “\(L\le 1\), so the series converges.” The Root Test gives absolute convergence only for \(L<1\), not for \(L=1\). The reciprocal-factor example shows the consequence: at \(x=2\), the limit is exactly \(1\), but the boundary series is harmonic and diverges.

Another source of errors is mishandling the factor outside the power while checking a particular term. For the first example at \(u=1\) and \(n=2\), the factor \(1/n^2\) is \(1/2^2=1/4\), not \(1/2\). The base is \(2/3\), so the term is \((1/4)(2/3)^2=1/9\). Keeping the prefactor and the powered base distinct prevents this arithmetic error.

Why the Boundary Determines the Final Parameter Set

The Root Test identifies an open region of parameters where the root limit is strictly below \(1\), and a region where it is strictly above \(1\). But the desired answer is often a complete set of parameter values, so leaving the boundary unresolved would leave the classification incomplete.

In the examples, the boundary series were respectively a reciprocal-square series, a harmonic series, and series whose terms fail to tend to zero. These different outcomes are compatible with the same Root Test boundary value \(L=1\). The exponential base determines the threshold; the remaining factors and signs can determine what happens exactly at that threshold.

Key takeaway: Solve the Root Test inequality using strict signs, then return to the original series at every equality case. A threshold is not itself a convergence conclusion.

Check Your Understanding

For each question, identify the strict Root Test cases and explain how you would handle any boundary values.

For \(\sum_{n=1}^{\infty}\frac{1}{n^3}\left(\frac{a}{4}\right)^n\), where \(a\ge 0\), what parameter range gives a root limit below \(1\)? What series appears at the threshold?
For \(\sum_{n=1}^{\infty}n\left(\frac{b+2}{3}\right)^n\), where \(b\) is real, solve the strict inequality for the Root Test convergence range. What must be checked at the two endpoints?
Why does a Root Test limit equal to \(1\) not prove either convergence or divergence?
For \(\sum_{n=1}^{\infty}\frac{1}{n^2}c^n\), what is the Root Test limit for fixed real \(c\), and which value of \(c\) requires separate examination?
At a parameter value where the base is \(-1\), why is it important to inspect the actual terms rather than conclude from the root limit alone?