Term-by-Term Operations on Power Series

Inside its interval of convergence, a power series behaves like a polynomial: you can differentiate it and integrate it term by term, and the radius of convergence \(R\) never changes. Add substitution and algebra, and one series — the geometric series — spawns a whole family of new ones.

By the end you'll be able to differentiate and integrate a power series term by term, build series for functions like \(\ln(1-x)\) and \(\frac{1}{(1-x)^2}\) from the geometric series, and say exactly what happens to the radius and the endpoints.

Predict: integrating the series term by term — does the interval's radius change? Toggle Integrate below to check, and watch the shaded convergence band on the number line.

Each bar is one term of the series at your chosen \(x\). Start with the geometric series \(\frac{1}{1-x} = \sum x^n\), then hit Differentiate (every bar becomes \(n\,x^{n-1}\), summing to \(\frac{1}{(1-x)^2}\)) or Integrate (every bar becomes \(\frac{x^{n+1}}{n+1}\), summing to \(-\ln(1-x)\)). The shaded band is where the series converges — its width never moves, but watch the endpoint dots switch between converge and diverge. Hover or tab to any bar for its exact value.

Operation

Sum of first 12 terms = 1.9995 · exact value 1/(1−x) = 2.0000 · R = 1 (unchanged)

1/(1−x) = Σ xⁿ, n ≥ 0 — first 12 terms at x = 0.50, and the convergence interval below
term values endpoint converges endpoint diverges

Within its interval of convergence a power series can be differentiated and integrated term by term — the radius \(R\) is unchanged — and known series can be manipulated algebraically to represent related functions such as \(\frac{1}{1-x}\) and its relatives.

Algebraic

Suppose \(f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n\) converges on \((a-R,\ a+R)\) — the interval whose radius \(R\) you found with the ratio test in the previous lesson. Then two things hold on that open interval. Differentiation: \(f'(x) = \sum_{n=1}^{\infty} n\,c_n (x-a)^{n-1}\) — apply the power rule to each term as if the sum were a finite polynomial. Integration: \(\int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1}\) — reverse power rule on each term, plus a constant of integration \(C\). In both cases the new series has the same radius \(R\), though endpoint behavior can change — recheck \(x = a \pm R\) separately.

Applied

In practice you rarely build a series from scratch. You start from a workhorse — the geometric series \(\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n\) for \(|x| < 1\) — and manipulate it. Substitute \(x \to -x\) to get \(\frac{1}{1+x} = \sum (-1)^n x^n\); multiply by \(x^2\) to get \(\frac{x^2}{1-x} = \sum x^{n+2}\); factor a constant, \(\frac{1}{5-x} = \frac{1}{5}\cdot\frac{1}{1-x/5} = \sum \frac{x^n}{5^{n+1}}\), now valid for \(|x| < 5\). Each move costs one line of algebra instead of a whole convergence analysis — but watch the convergence condition: substitution can move it, as \(|x/5| < 1\) becoming \(|x| < 5\) shows.

Worked example

Find a power series for \(\ln(1-x)\) and its interval of convergence. Start from \(\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n\) on \((-1,1)\). Since \(\int \frac{dx}{1-x} = -\ln(1-x) + C\), integrate the series term by term: \(-\ln(1-x) = C + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}\). Plug in \(x=0\): both sides give \(0\), so \(C = 0\). Multiply by \(-1\): \(\ln(1-x) = -\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots\right)\). Integration keeps \(R = 1\); checking endpoints, \(x=-1\) gives the alternating harmonic series (converges) and \(x=1\) gives the harmonic series (diverges), so the interval is \([-1, 1)\).

Your turn

Same play, different starting series: find a power series for \(\ln(1+x)\). Start from \(\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n\) (the \(x \to -x\) substitution). Since \(\int \frac{dx}{1+x} = \ln(1+x) + C\), integrating term by term gives \(\ln(1+x) = C + \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1}\). Now finish it: plug in \(x = 0\) to find \(C = \) ____, and state the series.

Reveal the answer

At \(x=0\): \(\ln(1) = 0\) and the sum is \(0\), so \(C = 0\). Therefore \(\ln(1+x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\), with radius \(R = 1\) preserved by the term-by-term integration. Forgetting \(C\) — or forgetting to solve for it at a convenient point — is the classic slip here.

More info — the substitution catalog, from another angle

Think of the geometric series as a template with a slot: \(\frac{1}{1-\square} = \sum \square^n\), valid whenever \(|\square| < 1\). Anything you can wedge into the slot yields a new series for free: \(\square = -x\) gives \(\frac{1}{1+x} = \sum (-1)^n x^n\); \(\square = x^2\) gives \(\frac{1}{1-x^2} = \sum x^{2n}\); \(\square = -x^3\) gives \(\frac{1}{1+x^3} = \sum (-1)^n x^{3n}\). The convergence condition travels with the slot — \(|x^2| < 1\) and \(|-x^3| < 1\) both still mean \(|x| < 1\), but \(|x/5| < 1\) means \(|x| < 5\) — so re-derive it every time rather than assuming it. This is the same building-new-from-old idea as the worked example above, just with substitution doing the work instead of integration; Paul's Online Math Notes (in Dive deeper below) drills exactly these moves.

Check your understanding

Question 1 of 4

Differentiate the geometric series \(\dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n\) term by term. What series do you get?

Question 2 of 4

Starting from \(\dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n\), substitute to find a power series for \(\dfrac{1}{1+x^3}\).

Question 3 of 4

You integrate a power series term by term inside its interval of convergence. Which of the following is guaranteed?

Question 4 of 4

A power series \(\sum_{n=0}^{\infty} c_n x^n\) has radius of convergence \(R = 3\), found with the ratio test. Its term-by-term derivative \(\sum_{n=1}^{\infty} n\,c_n x^{n-1}\) is guaranteed to converge for…

Recap

  • Term-by-term differentiation: \(f'(x) = \sum_{n=1}^{\infty} n\,c_n(x-a)^{n-1}\) — the power rule applied to each term, same radius \(R\).
  • Term-by-term integration: \(\int f\,dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1}\) — same radius \(R\); solve for \(C\) at a convenient point (usually \(x = a\)).
  • Only the radius is guaranteed — endpoints can change and must be rechecked (e.g. \(\sum x^n\) lives on \((-1,1)\) but its integral, the series for \(-\ln(1-x)\), lives on \([-1,1)\)).
  • Build new series from old: substitute into, multiply, and scale the geometric series \(\frac{1}{1-x} = \sum x^n\) to represent relatives like \(\frac{1}{1+x}\), \(\frac{1}{1-x^2}\), and \(\frac{1}{5-x}\) — re-deriving the convergence condition each time.

Dive deeper

Sources

  • Term-by-Term Differentiation, Integration, and Algebraic Manipulation of Power Series