Power Series and the Radius of Convergence
Read a power series as a function defined where it converges, find its radius with the ratio test, and check the endpoints separately.
What a learner can do afterwards
- Find the radius of convergence of a given power series
- Test both endpoints and state the full interval
- Differentiate or integrate a power series term by term inside its radius
1 · Read
A power series centered at c is a sum of terms a_n(x-c)^n, for n = 0, 1, 2, ... The coefficients a_n are fixed numbers. The n = 0 term is a_0, even at x = c. For finite R > 0, it converges absolutely when |x-c| < R and diverges when |x-c| > R. The boundaries need separate tests. R = 0 means only the center converges. R = infinity means the series converges for every real x.
Fix x and write u_n for the nth term. When consecutive ratios are eventually defined, find L = lim |u_(n+1)/u_n| as n grows. L < 1 proves absolute convergence. L > 1, including infinity, proves divergence. L = 1 gives no conclusion. Check the center directly, since its terms may make the ratio undefined. The ratio test does not settle every power series.
For terms (b(x-c))^n/n^p, with n >= 1, b nonzero and fixed p >= 0, the absolute ratio is |b(x-c)|(n/(n+1))^p away from c. Its limit is |b(x-c)|, giving R = 1/|b|. For p = 0 this is geometric: convergence holds exactly when |b(x-c)| < 1. At either boundary its terms have magnitude 1, so they do not tend to zero.
You have tested alternating series before. For sum (x+1)^n/(n*3^n), n >= 1, the ratio limit is |x+1|/3. Thus c = -1, R = 3 and the interior is -4 < x < 2. At -4 the terms are (-1)^n/n; the series converges by the alternating test. At 2 the terms are 1/n; the series diverges. The interval is [-4, 2). A square bracket includes a boundary; a parenthesis excludes it.
Here n! means the product 1*2*...*n. For terms n!(x-c)^n, n >= 1, the ratio is (n+1)|x-c| away from c. It tends to infinity, so only the center converges and R = 0. For terms (x-c)^n/n!, the ratio is |x-c|/(n+1). It tends to 0 for every fixed x away from c. The center also converges, so R = infinity.
Suppose the nonzero coefficients satisfy |a_(n+1)/a_n| -> A, where A is finite and positive. For sum a_n(x-c)^n, the term-ratio limit away from c is A|x-c|. Solve A|x-c| < 1 to get |x-c| < 1/A. Thus R = 1/A. Measure from the center to either boundary, not across the whole interval.
Differentiation finds a function's rate of change; f'(x) names its derivative. For a power series with R > 0, differentiate each term inside |x-c| < R: a_n(x-c)^n becomes n*a_n(x-c)^(n-1), for n >= 1. The constant term becomes 0. The differentiated series has the same radius. Check its endpoints again: terms that do not tend to zero prove divergence.
An antiderivative F of f satisfies F' = f. For continuous f, its definite integral from c to x is F(x)-F(c), which is 0 at x = c. Inside the radius, integrate a_n(x-c)^n to get a_n(x-c)^(n+1)/(n+1). Add C for an antiderivative; integrating from c to x gives C = 0 in this form. The new series keeps its radius; test its endpoints again. For a coefficient of power k, choose the original power k-1.
The geometric series with terms (x-1)^n/4^n, n >= 0, converges exactly when |x-1| < 4, so R = 4. Integrating from 1 to x gives terms (x-1)^(n+1)/((n+1)*4^n). Its radius remains 4. For instance, its quadratic term comes from n = 1 and is (x-1)^2/8.
Find the radius from a convergence test, check finite endpoints separately, and retain the radius after term-by-term differentiation or integration.
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More to read
Measuring from the center
A convergence interval can be shifted away from zero. Consider sum (x+5)^n/(n*2^n), starting at n = 1. Its center is -5. Away from that center, the absolute ratio of consecutive terms is |x+5|n/[2(n+1)]. As n grows, this tends to |x+5|/2.
The limit is below one when |x+5| < 2. This gives the interior -7 < x < -3 and radius 2. The width of that interior is 4. Radius measures from the center to a boundary, so it is half the width.
Substitute each boundary into the original terms. At x = -7, the terms are (-1)^n/n. Their magnitudes decrease to zero, so the alternating series converges. At x = -3, the terms are 1/n and the series diverges. The complete interval is [-7, -3). The radius calculation alone did not decide either boundary.
Changing the terms with calculus
Inside a power series' positive radius, its sum can be differentiated by differentiating each term. The constant disappears. For n >= 1, the term a_n(x-c)^n becomes n*a_n(x-c)^(n-1). The new series retains the original radius.
Endpoint convergence can change. For example, differentiating terms (x-c)^n/n cancels the denominator n and gives (x-c)^(n-1). At a boundary, inspect those new terms. If their magnitudes remain one, they cannot tend to zero, so the differentiated series diverges there.
Integration raises the power by one and divides by the new power. It sends a_n(x-c)^n to a_n(x-c)^(n+1)/(n+1). An antiderivative has an added constant C. Integrating from c to x makes that constant zero. The integrated series also retains the original radius. Test its boundary series separately before including either boundary.
Checking a radius calculation
Start by identifying the center and the first index. Evaluate the series at its center directly. This avoids dividing terms that might both be zero.
Away from the center, form the absolute ratio of consecutive terms when it is eventually defined. Then take its limit as the index grows. A limit below one proves absolute convergence. A limit above one proves divergence. A limit of one leaves the ratio test undecided.
The position of a factorial matters. For terms n!(x-c)^n, the ratio contains the growing factor n+1. Every fixed noncentral x gives divergence, leaving radius zero. For terms (x-c)^n/n!, the ratio instead contains 1/(n+1). Its limit is zero for each fixed noncentral x, and the center also converges. That series has infinite radius.
For a positive finite radius, find both boundaries by subtracting and adding the radius at the center. Substitute them into the original series. Record each convergence result before choosing brackets.
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