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Can you help me understand the highlighted parts of the proof. Thanks :)

Theorem: Let $\sum{a_nz^n}$ be a power series, let r be its radius of convergence. Then $\frac{1}{r} = \limsup |a_n|^{1/n}$.

Proof: Let $t = \limsup |a_n|^{1/n}$. Suppose first that $t \ne 0, \infty$. Give $\epsilon > 0$, there exits only a finite number of n such that $|a_n|^{1/n} \ge t + \epsilon$. Thus for all but a finite number of n, we have $|a_n| \le (t +\epsilon)^n$, whence the series $\sum{a_nz^n}$ converges absolutely if $|z| < \frac{1}{t+\epsilon}$, by comparison with the geometric series. Therefore the radius of convergence $r$ satisfies $r \ge \frac{1}{t+\epsilon}$ for every $\epsilon > 0$, whence $r\ge \frac{1}{t}$.

Conversely, given $\epsilon$ there exists infinitely many n such that $|a_n|^{1/n} \ge t-\epsilon$ and therefore, $|a_n|\ge (t-\epsilon)^n$. Hence the series $\sum{a_nz^n}$ does not converge if $|z| = \frac{1}{t-\epsilon}$, because it's $n^{th}$ term doesn't even tend to 0. Therefore $r \le \frac{1}{t-\epsilon}$ for every $\epsilon > 0$, whence $r\le \frac{1}{t}$. This concludes the proof for $t \ne 0,\infty$.


For the first highlighted part, I don't understand where the geometric series part comes from. $\sum{|a_n||z^n|} \le \sum{(1+\epsilon)^n\frac{1}{(t+\epsilon)^n}} = \sum1$, am I doing this wrong?

And on the second part why wouldn't the equality with $\frac{1}{t-\epsilon}$ make it converge?

Thanks!

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The Geometrie Series is $\sum_{n=1}^{\infty}q^n$ If your z is smaller than $\frac{1}{t–\epsilon}$ Then your q is smaller than 1 which means it Converges and behause it is a Majorat i.e. a bigger series which converges.

To your Second question if a Series converges than the nth term must converge to 0 because otherwise your series would not be cauchy anymore

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