A Practical Strategy for Finding a Radius
A power series radius calculation is often less about choosing between many complicated tests and more about noticing how the coefficients are built. A factorial or binomial coefficient usually simplifies in a quotient of consecutive coefficients. A polynomial factor usually contributes a quotient tending to \(1\). An exponential factor contributes a constant. Putting those pieces together often makes the radius immediate.
This workshop applies that strategy to several common combinations. The aim is to calculate the radius—not to determine what happens at the two boundary points. A radius calculation identifies the distance from the center within which the series converges and beyond which it diverges; it does not, by itself, classify convergence at the boundary.
For a fast calculation, first write down the coefficients \(a_n\), separate their different factors, and check which test turns those factors into manageable expressions. When consecutive coefficients are eventually nonzero, the ratio test is especially useful if \(\left|a_{n+1}/a_n\right|\) has a limit. The root test gives a general coefficient formula, including cases in which a consecutive-coefficient quotient is inconvenient.
Here is the reason the ratio formula works. Put \(u_n=a_n(x-c)^n\), the \(n\)th term of the series. For \(x\ne c\), and for sufficiently large \(n\) so that both coefficients are nonzero,
Thus, if the coefficient ratio tends to \(L\), the ratio of consecutive term magnitudes tends to \(L|x-c|\). The ratio test gives convergence when this limit is less than \(1\) and divergence when it is greater than \(1\). Those conditions are \(|x-c|<1/L\) and \(|x-c|>1/L\), respectively, which gives \(R=1/L\). At \(x=c\), the quotient above is not defined: every term with \(n\ge 1\) is zero. The series converges there directly, since only the constant term \(a_0\) can remain.
Worked Example: A Binomial Coefficient with an Exponential Factor
Worked Example: A Binomial Coefficient with an Exponential Factor
Find the radius of \(\sum_{n=0}^{\infty}\frac{\binom{2n}{n}}{9^n}(x+1)^n\). The center is \(c=-1\), and the coefficient is \(a_n=\binom{2n}{n}/9^n\). Since the coefficient contains a binomial coefficient, use the ratio test and simplify the quotient before taking a limit.
Using \(\binom{2n}{n}=(2n)!/(n!)^2\), calculate:
Therefore \(\lim_{n\to\infty}|a_{n+1}/a_n|=4/9\), because \((2n+1)/(n+1)\to 2\). The radius is
Check the calculation by substituting this value into the radius relation: \(1/R=1/(9/4)=4/9\), exactly the coefficient-ratio limit. Equivalently, for \(x\ne -1\), the term-magnitude ratio tends to \((4/9)|x+1|\); it is less than \(1\) when \(|x+1|<9/4\) and greater than \(1\) when \(|x+1|>9/4\). At \(x=-1\), the series converges directly, as only its constant term can be nonzero.
Factorial Ratios: Cancel Before Taking the Limit
When factorials appear in both numerator and denominator, expanding them into products is rarely the efficient first move. Instead, form \(a_{n+1}/a_n\). Most factorial factors cancel, leaving a small number of terms that reveal the limiting rate. Keep exponential factors visible while simplifying: they often supply the constant that determines the radius.
Worked Example: A Factorial Ratio and an Exponential Factor
Find the radius of \(\sum_{n=0}^{\infty}5^n\frac{(n!)^2}{(2n)!}(x-3)^n\). Here \(c=3\) and \(a_n=5^n(n!)^2/(2n)!\). The coefficients are nonzero, so the ratio test applies. Calculate the consecutive ratio:
The limit is \(5/4\): divide numerator and denominator of the last expression by \(n\) to obtain \(5(1+1/n)/(2(2+1/n))\to5/4\). Hence
As a check, \(1/R=1/(4/5)=5/4\), which agrees with the coefficient-ratio limit. For \(x\ne3\), the term-magnitude ratio tends to \((5/4)|x-3|\), giving the expected threshold \(|x-3|=4/5\). At \(x=3\), all terms after the constant term vanish, so the series converges there directly.
Polynomial Factors: Track Their Ratio, Not Their Size
A coefficient may grow polynomially as well as exponentially. For a polynomial \(P(n)\) with nonzero leading coefficient, the ratio \(P(n+1)/P(n)\) tends to \(1\), provided the ratio is defined for sufficiently large \(n\). Consequently, a polynomial factor changes the individual coefficients but does not change the limiting exponential rate. This conclusion is most directly visible in the ratio test.
Worked Example: A Polynomial Factor and an Exponential Factor
Find the radius of \(\sum_{n=0}^{\infty}(3n^2+2n+1)(2/5)^n(x-2)^n\). Its center is \(c=2\), and \(a_n=(3n^2+2n+1)(2/5)^n\). The polynomial factor is always positive for \(n\ge0\), so the coefficients are nonzero. Form their ratio:
The rational factor tends to \(1\), since its numerator and denominator have the same leading term \(3n^2\). Thus the ratio limit is \(2/5\), and the radius is
Check that the inverse radius is \(1/R=2/5\), matching the ratio limit. For \(x\ne2\), the term-magnitude ratio tends to \((2/5)|x-2|\), so the threshold distance is \(5/2\). At \(x=2\), the series reduces to its constant term and converges.
Combining Factors Without Losing the Rate
In a mixed coefficient, compute the ratio factor by factor. The quotient of a polynomial factor tends to \(1\); a binomial coefficient can be rewritten using factorials; and a factor \(q^n\) contributes \(q\). Multiplying these limiting contributions gives the ratio limit. This factor-by-factor approach is often clearer and less error-prone than estimating the size of the entire coefficient at once.
Worked Example: Binomial, Polynomial, and Exponential Factors Together
Find the radius of \(\sum_{n=0}^{\infty}(n+1)\binom{2n}{n}(2/15)^n(x-4)^n\). The center is \(c=4\), and the coefficient \(a_n=(n+1)\binom{2n}{n}(2/15)^n\) is positive. Its consecutive ratio has one factor from each part of the coefficient:
As \(n\to\infty\), \((n+2)(2n+1)/(n+1)^2\to2\), so the ratio limit is \(8/15\). Therefore,
Check: \(1/R=8/15\), precisely the ratio limit. For \(x\ne4\), the term-magnitude ratio tends to \((8/15)|x-4|\), which is less than \(1\) inside distance \(15/8\) and greater than \(1\) outside it. At \(x=4\), the series converges directly because only the constant term remains.
Choosing Between the Ratio and Root Tests
The examples use the ratio test because consecutive coefficient quotients simplify cleanly. A practical choice of test depends on the form of the coefficients:
| Coefficient feature | Efficient first approach |
|---|---|
| Factorials or binomial coefficients | Form \(a_{n+1}/a_n\) and cancel factorials. |
| A polynomial in \(n\) | Take its consecutive quotient; for a nonzero polynomial it tends to \(1\). |
| An exponential factor \(q^n\) | Keep track of its exact contribution, \(q\), to the ratio. |
| A simple expression for \(|a_n|^{1/n}\), but an awkward consecutive quotient | Use the root-test formula for the radius. |
The root-test formula is useful when taking the \(n\)th root makes the coefficient structure simpler. For example, factors that are fixed constants raised to the \(n\)th power become constants after taking an \(n\)th root. The formula also avoids requiring every consecutive quotient to be defined. In either approach, calculate the limiting rate first, then take its reciprocal to obtain a finite positive radius.
Common Pitfall: Dividing by Zero at the Center
A ratio of consecutive terms is a tool for applying the ratio test, but it must be defined before it is used. For \(u_n=a_n(x-c)^n\), it is incorrect to write the quotient identity at \(x=c\): if \(n\ge1\), then \(u_n=0\), so the quotient of consecutive terms has the form \(0/0\). Having nonzero coefficients \(a_n\) does not fix this problem. The identity is valid only when \(x\ne c\) and the relevant coefficients are nonzero.
For instance, if \(a_n=1\), a mistaken calculation at the center \(c=2\) might claim that \(\left|u_{n+1}/u_n\right|=|x-2|=0\), then conclude that the ratio tends to zero. At \(x=2\), however, the actual quotient is \(0/0\), not zero. The correction is to treat the center directly: the series \(\sum_{n=0}^{\infty}(x-2)^n\) has value \(1\) at \(x=2\), because its constant term is \(1\) and every later term is zero.
Check Your Understanding
For each question, choose a suitable coefficient test and identify the calculation that determines the radius. Do not classify boundary convergence.
- For \(a_n=\binom{2n}{n}/16^n\), what is the limit of \(|a_{n+1}/a_n|\), and what radius follows?
- For \(a_n=3^n(n!)^2/(2n)!\), simplify \(|a_{n+1}/a_n|\) before taking its limit.
- Why does multiplying coefficients by a nonzero polynomial in \(n\) usually leave the limiting coefficient ratio unchanged?
- For \(a_n=(n+2)(4/7)^n\), calculate the ratio limit and the corresponding radius.
- Why must the center of a power series be handled separately when using a quotient of consecutive terms?
