Root Test Proof Verification

Question

Prove the following: (Root Test) Given a sequence $a_n$ of complex numbers, let $$A=\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}$$

i)If $A<1$, the series $\sum a_n$ converges.

ii)If $A>1$, the series $\sum a_n$ does not converges.

My Proof:

i) If $A<1$:

Choose a positive $E$ such that $A+E<1$,

Since $A=\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}$, there exists a natural number $m$ such that: $|a_n|^{1/n}<A+E$ for all $n\geq m$.

let $A+E=P$, then $0<P<1$ and $|a_n|<P^n, \forall n\geq m$.

But, $\sum P^m$ is a convergent series since $0<P<1$ .

By comparison test, the series $\sum a_n$ is convergent.

ii) If $A>1$:

Choose a positive $E$ such that $A-E>1$,

Since $A=\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}$, $|a_n|^{1/n}>A-E$ for infinite number of values of $n$.

That is, infinite number of elements of the sequence $a_n$ are greater than $1$ and therefore $A=\limsup_{n\rightarrow\infty}|a_n|\neq 0$.

Hence, if $A>1$, the series $\sum a_n$ is not convergent.

Answer 1

The proof is just fine. There is a slight problem with the language: where you wrote “infinite number of elements of the sequence”, it would be better to write “infinitely many elements of the sequence”.

Answer 2

Perhaps, $|a_{n_{k}}|>(A-E)^{n_{k}}$, where $(n_{k})$ is strictly increasing, then $\limsup_{n\rightarrow\infty}|a_{n}|\geq\limsup_{k\rightarrow\infty}|a_{n_{k}}|\geq\lim_{k\rightarrow\infty}(A-E)^{n_{k}}=\infty$, so $\lim_{n\rightarrow\infty}a_{n}=0$ does not hold and hence the series does not converge.