1. ## Abel's Limit Theorem

Theorem. If $\displaystyle \sum_{0}^{n} a_{n}$ converges then $\displaystyle f(z) = \sum_{0}^{\infty} a_{n}z^{n}$ tends to $\displaystyle f(1)$ as $\displaystyle z$ approaches $\displaystyle 1$ in such a way that $\displaystyle |1-z|/(1-|z|)$ remains bounded.

Here was assuming that $\displaystyle R=1$ and convergence takes place at $\displaystyle z = 1$.

But we let $\displaystyle R = k$ and $\displaystyle z = k$ for example, right?

2. Originally Posted by manjohn12
Theorem. If $\displaystyle \sum_{0}^{n} a_{n}$ converges then $\displaystyle f(z) = \sum_{0}^{\infty} a_{n}z^{n}$ tends to $\displaystyle f(1)$ as $\displaystyle z$ approaches $\displaystyle 1$ in such a way that $\displaystyle |1-z|/(1-|z|)$ remains bounded.

Here was assuming that $\displaystyle R=1$ and convergence takes place at $\displaystyle z = 1$.

But we let $\displaystyle R = k$ and $\displaystyle z = k$ for example, right?
That is correct. The usual statement of Abel's theorem is that if the series $\displaystyle f(z) = \sum_{0}^{\infty} a_{n}z^{n}$ has radius of convergence R, and converges at a point $\displaystyle z_0$ on the circle of convergence (so $\displaystyle |z_0|=R$), then $\displaystyle \lim f(z) = f(z_0)$ as $\displaystyle z_0\to z$ nontangentially (meaning that that the angle between $\displaystyle z_0-z$ and the tangent at $\displaystyle z_0$ is bounded away from 0).

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# abel's limit theorem state

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