The Root Test given in my university textbook is as follows :
Let $\displaystyle L = \limsup_{n \to ∞} \sqrt[n]{|a_n|}$. If $L<1$ then the series is convergent. If $L>1$ then the series is divergent. If $L=1$ then the test fails. Above is the background info. However, I would like to discuss further on the distinction between
- case 1: $1<L<+∞$ and
- case 2: $L=+∞$.
Although the series diverge in both cases, they seem quite different for me, as one grows significantly faster than the other. Is there a way to make use of this distinction?
Possible trains of thoughts:
1.take the sequence of partial sums and form a series out of this sequence
2.use the sequence of partial sums as the coefficients of a power series.
Forget about the three cases, let's focus on the proof of the root test.
$\displaystyle L=\lim\sup_{n\to\infty}\sqrt[n]{|a_n|}$ implies there is a subsequence $\{a_{n_k}\}$ of $\{a_n\}$ such that
$\displaystyle\lim_{n_k\to\infty}\sqrt[n_k]{|a_{n_k}|}=L\in [0,\infty]$.
If $L<\infty$, then for every $\varepsilon>0$, there exists $N>0$, such that for every $n>N$, $\displaystyle \sqrt[n]{|a_n|}<L+\varepsilon$.
If $L<1$, then by 2. we know that $|a_n|<(L+\varepsilon)^n$ for sufficiently large $n\ge N$ and we can take $\varepsilon$ so that $L+\varepsilon<1$. Then $$\sum_{n=N}^\infty|a_n|<\sum_{n=N}^\infty (L+\varepsilon)^n<\infty,\quad \text{since}\ (L+\varepsilon)<1.$$
If $1<L<\infty$, then there by 1. we know that $|a_{n_k}|>(L-\varepsilon)^{n_k}$ for some $n_k$ sufficiently large, where $\varepsilon$ is taken so that $L-\epsilon>1$. Since $a_{n_k}$ doesn't converge to $0$, then of course $\{a_n\}$ diverges.
Now can you write down the proof for $L=\infty$?