# Understanding Gaps in concept of Radius of Convergence of Power Series

I had done this topic Many time Thinking it is easy But When I today done its theorem , I come across that I had certain gaps in My Knowelege about it.

$$\sum a_nz^n$$ is power series. $$1/R=limsup (a_n)^{1/n}$$
1) $$|Z| ,then series converges absolutely
2) $$|z|>R$$, then series diverges

Let $$L=limsup (a_n)^{1/n}$$
By defination $$\forall \epsilon >0 \exists n_1\in N$$ such $$(a_n)^{1/n} Now $$a_n<(L+\epsilon )^n$$ i.e $$a_n<(1/R+\epsilon )^n$$

$$|\sum a_nz^n|\leq \sum (1/R+\epsilon )^nz^n$$
$$(z/R+\epsilon z)^n$$<1
[By 1 $$|z|/R<1$$ and for fix z we can vary any $$\epsilon$$ we wanted ]
So by Root test , RHS series is convergent
SO Original power series also converges

Similary for 2.
Is there is any gaps in my argument ?
Any Help will be appreciated

• Your argument looks good, except you will want absolute values on some of the terms. For example, $R=\lim\sup |a_n|^{1/n}$ rather than $\lim\sup a_n^{1/n}$, and your sum should be $$|\sum a_nz^n|\leqslant \sum |(1/R+\epsilon)^n z^n|.$$ Also, you only need $|z/R+\epsilon z|<1$ rather than $|z/R+\epsilon z|^n<1$. Then you can use comparison tests to the geometric series $sum |z/R+\epsilon z|^n$ to deduce convergence (which is what root test does anyway). – bangs Sep 21 '18 at 12:21
• Choosing the $n$th root test for convergence is natural given the hypothesis on power series coefficients $a_n$, but this test should be stated, so all details can be verified. Then part 1 can be completed. Part 2 is about divergence, so saying it can be done similarly is not clear. Perhaps stating the root test rigorously will make the similarity evident? – hardmath Sep 21 '18 at 17:33