Convergence of Power Series¶

Theorem¶

Consider a power series $\displaystyle f(x)=\sum_{n=0}^\infty c_n (x-a)^n$ with a radius of convergence $R > 0\text{.}$ For any chosen $r$ such that $0 < r < R$$R$:

• The series converges pointwise on the open interval $(a-R, a+R)\text{.}$
• It converges uniformly on the closed interval $[a-r, a+r]\text{.}$
• The limit function $f$ is differentiable on $(a-R, a+R)\text{.}$
• The derivative series $\displaystyle \sum_{n=1}^\infty nc_n (x-a)^{n-1}$ converges uniformly to the derivative, $f'\text{,}$ on $[a-r, a+r]\text{.}$

The smaller closed Interval $[a-r,a+r]$ stays strictly inside the convergence interval, which is why one uniform tail estimate works there; the endpoints $a\pm R$ need separate tests.

Interpretation¶

Inside its radius of convergence a power series can be handled almost like a polynomial: it may be differentiated term by term, and the differentiated series converges to the derivative of the sum. The price is that uniform convergence is only guaranteed on closed intervals strictly inside $(a-R,a+R)$$R)$.

The open interval $(a-R,a+R)$ is where the series converges pointwise; the closed interval $[a-r,a+r]$$r]$, with $0<r<R$$R$, is where the convergence is uniform. The gap between $r$ and $R$ cannot be removed in general: the geometric series $\displaystyle \sum_n x^n$ has $a=0$ and $R=1$$1$, but it does not converge uniformly on $(-1,1)$, since its remainder $x^{N+1}/(1-x)$ is unbounded as $x\to1^-$$1^{-}$.

The uniform convergence on $[a-r,a+r]$ comes from the Weierstrass $M$-test: for $|x-a|\le r$ we have $|c_n(x-a)^n|\le|c_n|r^n$$|c_n|r^n$, and $\displaystyle \sum_n|c_n|r^n$ converges because $r<R$$R$. The derivative series has the same radius $R$, so the same argument applies to it, and the termwise differentiation result for an infinite series of functions gives $f'$. Every $x$ in $(a-R,a+R)$ lies in some interval $[a-r,a+r]$$r]$, so $f$ is differentiable at every point of the open interval although the convergence need not be uniform on all of it.

The theorem says nothing about the endpoints $a\pm R$$R$. There the series and its derivative series can behave differently: $\displaystyle \sum_{n\ge1}x^n/n^2$ converges at $x=1$$1$, while its derivative series $\displaystyle \sum_{n\ge1}x^{n-1}/n$ is the harmonic series at $x=1$ and diverges.

Repeating the argument shows that $f$ is infinitely differentiable on $(a-R,a+R)$$R)$, and differentiating $n$ times at $a$ gives $c_n=f^{(n)}(a)/n!$$f^{(n)}(a)/n!$. This formula for the coefficients is the starting point of the Taylor series.

Key Results¶

• Term-by-term differentiation is valid inside the radius of convergence.
• Locally (on $[a-r,a+r]$$r]$), a power series behaves like a polynomial: it is smooth and uniformly controlled.

Examples¶

• The series $\displaystyle \sum_{n=0}^\infty n(-2)^n (x-3)^n$ with $R = 1/2$ centred at $a = 3\text{,}$ converges pointwise on $(5/2, 7/2)$ and uniformly on any closed interval within it.
• The series $\displaystyle \sum_{n=0}^\infty 2^{n^2} (x+2)^n$ with $R = 0$ centred at $a = -2\text{,}$ converges solely at its centre $x = -2\text{.}$
• The series $\displaystyle \sum_{n=0}^\infty 2^{-n^2} (x+2)^n$ with $R = +\infty$ centred at $a = -2\text{,}$ converges pointwise over the entire real line $\mathbb{R}\text{.}$ It converges uniformly on every closed interval $[-2-r,-2+r]$$r]$, but not on $\mathbb{R}$: for $n\ge1$ the term $2^{-n^2}(x+2)^n$ is unbounded on $\mathbb{R}$, while uniform convergence would force the terms to tend to $0$ uniformly.

Prerequisites¶

• Interval
• Radius and Interval of Convergence

Related Concepts¶

• Taylor Series
• Power Series

Sources¶

• Math III: Analysis · Lecture 2

Symbols: infinity, real numbers, sum

Last updated: 20 October 2026 03:02