Operating on Power Series

A power series can be manipulated inside its interval

Within its interval of convergence, a power series behaves like a polynomial in ways that are remarkably useful. It can be added to another series, multiplied by a constant or by a power of $x$, differentiated term by term, and integrated term by term. These operations generate representations for functions that may not have been given as a series initially. The crucial qualification is the interval: an algebraically correct new expression still needs a convergence statement.

Start from the geometric series

$$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n =1+x+x^2+x^3+\cdots,\qquad |x|<1.$$

The equality is not true for every real $x$. At $x=1$, the right side is $1+1+1+\cdots$, which diverges; at $x=-1$, it alternates without converging. The open interval $(-1,1)$ is part of the statement, not a footnote. It tells us where the infinite polynomial actually represents the rational function.

Differentiate coefficient by coefficient

Differentiating the geometric series inside $|x|<1$ gives

$$\frac{1}{(1-x)^2} =\sum_{n=1}^{\infty}nx^{n-1} =1+2x+3x^2+4x^3+\cdots.$$

Each term uses the ordinary power rule. The derivative of $x^n$ is $nx^{n-1}$, and the constant term differentiates to zero. Reindexing is optional but can make a target coefficient easier to read. For example, replacing $n-1$ by $k$ yields $\sum_{k=0}^{\infty}(k+1)x^k$, which begins at a constant power of $x$.

The radius of convergence stays the same under differentiation, but endpoint behavior can change. The original geometric series diverges at both endpoints. Its derivative series also diverges at $x=1$, and at $x=-1$ it becomes $1-2+3-4+\cdots$, which still diverges because the terms do not approach zero. This example does not establish a universal endpoint rule; it demonstrates why endpoints must be retested after an operation.

The reason term-by-term differentiation is powerful is that it converts a known representation into a new one without re-deriving the function from scratch. It is not permission to differentiate a divergent numerical series at an endpoint. The theorem applies throughout the interior of the radius, where the power series has the needed convergence behavior. A final representation should therefore name both the derived function and the interval on which the equality has been established.

Integrate coefficient by coefficient

Integrating from $0$ to $x$ is often the cleanest way to control the constant of integration:

$$\int_0^x\frac{1}{1-t}\,dt =\int_0^x\sum_{n=0}^{\infty}t^n\,dt =\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}.$$

The left side is $-\ln(1-x)$, so

$$-\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n},\qquad -1\le x<1.$$

The interval changed at the left endpoint. Substituting $x=-1$ produces the alternating harmonic series, which converges, while substituting $x=1$ produces the harmonic series, which diverges. Definite integration automatically selects the constant because both sides equal zero at $x=0$. If indefinite integration is used instead, the constant must be determined from a known value.

Multiply by powers and combine series

Multiplying a series by $x^m$ shifts every exponent by $m$. Thus

$$\frac{x^3}{1-x}=x^3+x^4+x^5+\cdots$$

for $|x|<1$. The shift is visible in the first terms and should be written before applying a summation formula. Multiplying two full series requires collecting like powers. The coefficient of $x^n$ in a product is formed from every pair of terms whose exponents add to $n$; this is called a Cauchy product. It is not generally obtained by multiplying coefficients with the same index only.

Addition and subtraction are simpler: align powers and combine their coefficients. If one series begins at $n=1$ and another at $n=0$, write a few terms rather than forcing them to share an index immediately. Index notation is concise only after it faithfully represents the visible pattern. In every operation, the new series is certainly valid on the overlap of the original intervals; a separate test may reveal a larger endpoint inclusion, but it should not be assumed.

This is a good place to distinguish equality of functions from equality of formal-looking symbols. On a shared interval, two power-series expressions may represent the same function and can be combined point by point. Outside that interval, one or both series may fail even if a simplified algebraic formula is defined there. The rational expression $1/(1-x)$, for example, is defined at $x=2$, but its geometric series is not. A series representation carries more information than the closed-form function alone.

Retrieval check

Beginning with $1/(1-x)=\sum_{n=0}^{\infty}x^n$, derive a power series for $1/(1-x)^2$ and then for $\ln(1-x)$. Include the interval after each step and test both endpoints of the logarithm series. Finally, explain why integrating a series does not authorize carrying the original endpoint brackets forward without checking. The next lesson builds Taylor polynomials by matching derivatives at a chosen center, which gives a systematic way to construct local power-series approximations.