Power series

A power series is an infinite polynomial, $a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots$ , and the single most useful fact about it is that it does not converge everywhere: there is a radius $R$ around its center $c$ inside which the series behaves like a perfectly ordinary function, and outside which it simply blows up. Power series is the capstone of the Unit 4 chapter "Sequences, Series, and Power Series" of MA10001 — it reuses the ratio and root tests built earlier in the chapter to pin down $R$ , and the comparison and alternating-series tests to settle the two endpoints. It is also the direct bridge to Unit 1's Taylor and Maclaurin theorem, since writing a function as its Taylor series is exactly writing it as one particular power series, and it looks forward to Unit 5's complex Taylor and Laurent series, where the same radius-of-convergence machinery reappears unchanged. This article covers finding $R$ and the full interval of convergence, differentiating and integrating a power series term by term, and building new power series from the standard Maclaurin series for $e^x$ , $\sin x$ , $\cos x$ , and $\ln(1+x)$ .

The core insight

Every power series has a radius of convergence $R$ found from its coefficients by the ratio or root test, converges absolutely inside $|x-c|<R$ , diverges outside it, and can be differentiated or integrated term by term anywhere inside that same radius.

The core idea

A power series centered at $c$ is the series

\sum_{n=0}^{\infty} a_n (x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots

where: $a_n$ are the coefficients (fixed numbers, independent of $x$ ), $c$ is the center, and $x$ is the variable. Setting $x=c$ always gives the single term $a_0$ , so a power series converges trivially at its own center; the interesting question is how far from $c$ it keeps converging.

Radius of convergence. There is a number $R \in [0,\infty]$ , computed from the coefficients alone, such that the series converges absolutely for $|x-c|<R$ and diverges for $|x-c|>R$ . The two boundary points $|x-c|=R$ are not decided by this test and must be checked one at a time:

\frac{1}{R} = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| \qquad \text{or, equivalently,} \qquad \frac{1}{R} = \limsup_{n\to\infty} |a_n|^{1/n}.

where: the left-hand formula is the ratio test applied to $|a_n|$ (not to the full term $a_n(x-c)^n$ ), and the right-hand formula is the root test; either gives the same $R$ , and a problem usually favors whichever makes the algebra cleaner (factorials favor the ratio test, $n$ -th powers favor the root test). $R=0$ means the series converges only at $x=c$ ; $R=\infty$ means it converges for every real $x$ .

Interval of convergence is the open interval $(c-R, c+R)$ from the radius test, adjusted by including whichever of the two endpoints $x = c-R$ and $x = c+R$ turn out, on separate inspection, to converge.

Term-by-term differentiation and integration. If $f(x) = \sum_{n=0}^\infty a_n(x-c)^n$ has radius of convergence $R$ , then for every $x$ strictly inside $(c-R,c+R)$ ,

f'(x) = \sum_{n=1}^{\infty} n\,a_n (x-c)^{n-1}, \qquad \int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{a_n}{n+1}(x-c)^{n+1}.

where: the differentiated series is obtained by the ordinary power rule applied to each term, and the integrated series by the ordinary power rule for integration; both new series have exactly the same radius of convergence $R$ as the original, though the endpoint behavior at $x=c\pm R$ can change (a series can gain or lose convergence at an endpoint after differentiating or integrating, even though $R$ itself never changes). The figure below shows the shape of this picture on the number line: absolute convergence fills the open interval, divergence fills everything outside it, and the two endpoints are the only points that need individual attention.

Figure

A number line showing absolute convergence inside radius R around center c, divergence outside, and the two endpoints marked for separate testing

Finally, the four standard Maclaurin series (power series about $c=0$ ) below are all direct consequences of Unit 1's Taylor's theorem, and every one is used constantly for building new series by substitution, differentiation, or integration:

e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \quad \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}, \quad \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n}.

where: the series for $e^x, \sin x, \cos x$ all have $R=\infty$ (they converge for every real $x$ ), while the series for $\ln(1+x)$ has $R=1$ , converging on $-1 < x \le 1$ .

Derivation ARadius and interval of convergence from the ratio test

The goal is to show, on a fully general power series $\sum a_n(x-c)^n$ , exactly how the ratio test on the coefficients produces both the radius $R$ and, after a separate endpoint check, the full interval of convergence — this is the one procedure every convergence question in this chapter reduces to, and it is worth walking through once in full generality before touching numbers.

Step 1. Form the ratio of consecutive terms of the series (not just the coefficients) and take its limit as $n\to\infty$ : $\displaystyle L(x) = \lim_{n\to\infty}\left|\frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n}\right| = |x-c|\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$ . The factor $|x-c|$ pulls straight out because it does not depend on $n$ , leaving a pure limit of the coefficients alone.

Step 2. By the ratio test for ordinary series, $\sum a_n(x-c)^n$ converges absolutely wherever $L(x) < 1$ and diverges wherever $L(x) > 1$ . Writing $\ell = \lim|a_{n+1}/a_n|$ , the convergence condition $L(x) = |x-c|\,\ell < 1$ rearranges to $|x-c| < 1/\ell$ . That threshold $1/\ell$ is exactly the radius of convergence: $R = 1/\ell$ .

Step 3. This pins down the open interval $(c-R, c+R)$ where the series converges absolutely, and confirms divergence for $|x-c|>R$ , but the ratio test gives no information at $L(x)=1$ , which is precisely $|x-c|=R$ — the two endpoints $x=c-R$ and $x=c+R$ .

Step 4. Resolve each endpoint by substituting it into the original series and applying an earlier test (comparison, limit comparison, or the alternating-series/Leibniz test) to that specific numerical series. The two endpoints must be checked independently — a series can converge at one and diverge at the other — and only after both checks is the full interval of convergence (open, half-open, or closed) known.

Worked example

Find the radius and full interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x-2)^n}{n\,3^n}$ .

Step 1. Apply the ratio test to the coefficients $a_n = \dfrac{1}{n\,3^n}$ of $(x-2)^n$ :

\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{1}{(n+1)3^{n+1}} \cdot \frac{n\,3^n}{1}\right| = \frac{n}{n+1}\cdot\frac{1}{3} \longrightarrow \frac{1}{3} \quad \text{as } n\to\infty.

So $\ell = 1/3$ , giving $R = 1/\ell = 3$ by Step 2 of the derivation above.

Step 2. The open interval of convergence is $(2-3,\ 2+3) = (-1,5)$ . Now check the two endpoints separately, exactly as the derivation requires. At $x=5$ : the series becomes $\sum \dfrac{3^n}{n\,3^n} = \sum \dfrac{1}{n}$ , the harmonic series — this diverges. At $x=-1$ : the series becomes $\sum \dfrac{(-3)^n}{n\,3^n} = \sum \dfrac{(-1)^n}{n}$ , the alternating harmonic series — by the Leibniz test this converges (conditionally, not absolutely).

Step 3. Including the convergent endpoint and excluding the divergent one, the full interval of convergence is

[-1,\ 5).

Competitive stretch

A related skill this chapter tests is building a new power series from a known one by substitution and term-by-term integration, staying strictly inside the radius of convergence. Take the Maclaurin series $e^u = \sum_{n=0}^\infty u^n/n!$ , valid for every real $u$ ( $R=\infty$ ), and substitute $u=-t^2$ :

e^{-t^2} = \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!}, \qquad \text{valid for all real } t.

Because this series has infinite radius of convergence, term-by-term integration is valid on any interval $[0,x]$ , giving

\int_0^x e^{-t^2}\,dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\int_0^x t^{2n}\,dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\cdot\frac{x^{2n+1}}{2n+1}.

This is a convergent power series, for every real $x$ , representing the integral $\int_0^x e^{-t^2}dt$ — a function with no elementary closed form — obtained purely by substitution into a standard series followed by legal term-by-term integration.

Common traps

Forgetting to check the endpoints separately. The ratio/root test only pins down the open interval $(c-R,c+R)$ ; behavior exactly at $x=c\pm R$ is never decided by these tests and must be checked one endpoint at a time.
Assuming endpoint behavior is symmetric. The worked example converges at $x=-1$ but diverges at $x=5$ — there is no shortcut; each endpoint needs its own test.
Applying term-by-term differentiation/integration outside the radius of convergence. These operations are guaranteed valid only strictly inside $(c-R, c+R)$ ; at the endpoints the differentiated/integrated series can behave differently from the original.
Mixing up the radius formula. $1/R = \lim|a_{n+1}/a_n|$ uses the coefficients $a_n$ alone, not the full term $a_n(x-c)^n$ ; plugging in the full term is fine as a starting point (as in Step 1 of the derivation) but the $|x-c|$ factor must be isolated correctly before comparing to $1$ .

Ask the tutor: Want to work through finding the radius of convergence of a trickier power series, or see how substitution and term-by-term integration turn one known Maclaurin series into another? Ask the AI tutor to walk through it with you.

Why does the ratio test give R = 1/l, and not just l itself?

The ratio test says the series converges when $|x-c|\cdot \ell < 1$ , where $\ell = \lim|a_{n+1}/a_n|$ . Solving that inequality for $|x-c|$ gives $|x-c| < 1/\ell$ , so the radius is the reciprocal of $\ell$ , not $\ell$ itself. It is easy to forget the flip under a time crunch — a quick sanity check is that a larger $\ell$ (coefficients shrinking more slowly) should mean a smaller radius, since $1/\ell$ then gets smaller, which matches intuition that slower-shrinking coefficients tolerate less growth in $|x-c|^n$ .

If the radius of convergence is R=infinity, do I still need to check endpoints?

No. Endpoints only exist to check when $R$ is finite, because $R=\infty$ means the interval of convergence is already the entire real line $(-\infty,\infty)$ with no boundary points left over. This is exactly the situation for $e^x$ , $\sin x$ , and $\cos x$ above — their ratio tests give $\ell=0$ , so $R=1/0=\infty$ , and there is nothing more to check.

Can I substitute any expression for x into a known Maclaurin series, like in the worked example's stretch?

Yes, but the resulting series is only valid where the substituted expression itself lies inside the original series' radius of convergence. For $e^u=\sum u^n/n!$ the radius is infinite, so any real substitution $u=-t^2$ stays valid for all real $t$ , as used above. If instead you substituted into $\ln(1+u)=\sum(-1)^{n-1}u^n/n$ , which only has $R=1$ , you would need $|u|<1$ (with the endpoint $u=1$ also valid) before the substituted series is trustworthy.

Power series

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads