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.
Please Help me to fill that

$\sum a_nz^n$ is power series. $1/R=limsup (a_n)^{1/n}$
1) $|Z|<R $ ,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}<L+\epsilon $ 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

  • $\begingroup$ 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). $\endgroup$ – bangs Sep 21 '18 at 12:21
  • $\begingroup$ 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? $\endgroup$ – hardmath Sep 21 '18 at 17:33

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