Abel's Limit Theorem

**manjohn12** · March 26th 2009, 10:46 PM

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

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

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

**Opalg** · March 29th 2009, 11:52 AM

Originally Posted by manjohn12

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

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

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

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

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