Theorem:6481641

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theorem

Named after: Niels Henrik Abel

This page was built for theorem: Abel's theorem

Theorem

Suppose that $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ has a radius of convergence $r$ and that $\sum _{n=0}^{\mathrm{\infty }}{a}_n{r}^n$ is convergent. Then $$\underset{x\to r^{-}}{\lim }\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=\sum _{n=0}^{\mathrm{\infty }}{a}_n{r}^n.$$

Proof

Suppose that $$\sum _{n=0}^{\mathrm{\infty }}{a}_n=L$$ is a convergent series, and define $$f(r)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{r}^n$$ The convergence of the first series implies that $a_n\to 0$, and hence $f(r)$ converges for $|r|<1$. We will show that $f(r)\to L$ as $r\to 1^{-}$.

Let $$s_N=a_0+a_1+\cdots +a_N,N\in \mathbb{N}$$ denote the corresponding partial sums. Our proof relies on the following identity: $$f(r)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{r}^n=(1-r)\sum _{n=0}^{\mathrm{\infty }}{s}_n{r}^n.$$ (1)

The above identity obviously works at the level of formal power series. Indeed, $$a_0+(a_1+a_0)r+(a_2+a_1+a_0)r^2+\cdots -(a_{0}r+(a_1+a_0)r^2+\cdots )=a_0+a_{1}r+a_2{r}^2+\cdots$$

Since the partial sums $s_n$ converge to $L$, they are bounded, and hence $$\sum _{n=0}^{\mathrm{\infty }}{s}_n{r}^n$$ converges for $|r|<1$. Hence, for $|r|<1$, identity (1) is also a genuine functional equality.

Let $ϵ>0$ be given. Choose an $N$ sufficiently large so that all partial sums $s_n$ with $n>N$ satisfy $$|s_n-L|\le ϵ$$ Then, for all $r$ such that $0<r<1$, one obtains $$|\sum _{n=N+1}^{\mathrm{\infty }}(s_n-L)r^n|\le \frac{ϵr^{N+1}}{1-r}$$

Note that $$f(r)-L=(1-r)\sum _{n=0}^N(s_n-L)r^n+(1-r)\sum _{n=N+1}^{\mathrm{\infty }}(s_n-L)r^n$$

As $r\to 1^{-}$, the first term tends to 0. The absolute value of the second term is estimated by $$ϵr^{N+1}\le ϵ$$ Hence, $$\underset{r\to 1^{-}}{\mathrm{lim\; sup}}|f(r)-L|\le ϵ$$

Since $ϵ>0$ was arbitrary, it follows that $f(r)\to L$ as $r\to 1^{-}$.

Lean Prover

Abel's Theorem

Sources

Abel’s limit theorem in Mathplanet

Proof of Abel's Convergence Theorem in Mathplanet