Radius of Convergence

maths ✒ radius-of-convergence

Given a power series $ \Sigma a_n z^n $,
the radius of convergence $ R = sup \{ ~ z ~ | ~ \Sigma ~ a_n z^n ~ \mbox{convergent} ~ \} $.

If $ \Sigma ~ a_n z^n ~ \mbox{convergent} ~ ~ \forall z $, then we write $ R = \infty $.

Convergent within the radius of convergence

If $ |z| < |y| $ and $ a_n y^n ~ \mbox{convergent} $ then $ a_n z^n ~ \mbox{convergent} $.

Choose p such that $ \frac {|x|} {|y|} < p < 1 $.

As $ a_n z^n ~ \mbox{convergent} $, $ a_n z^n \to 0 $ so
$ \exists N \mbox{s.t.} \forall n \ge N, ~ a_n z^n < 1 $.

Then $ \forall n \ge N, | a_n y^n | = | a_n z^n | {\left | \frac {y^n} {z^n} \right |} < p^n $.

Then by the Comparison Test on the series $ p^N + p^{N+1} + ... $ the N-tail of $ a_n y^n $ is absolutely convergent.

Therefore $ a_n y^n $ is convergent.

If $ |z| < R $ then $ a_n z^n ~ \mbox{convergent} $.

By the definition of the supremum, there $ \exists y \in (|z|, R) $ such that the power series is convergent at y.

Then apply the above argument to z and y.