I have been trying my own proof of the root test since I don't quite understand this one.
So I will show what I have and explain why I'm stuck.
Suppose $a_n \geq 0$ and $\displaystyle{\lim_{n\to \infty}} \sqrt[n]{a_n} = r$.
Claim: $\sum_{n=1}^\infty a_n$ converges if $r < 1$.
Proof: We know that $r \geq 0$ since $\sqrt[n]{a_n} \geq 0$, so suppose $0 \leq r < 1$.
Then $\displaystyle{\lim_{n\to 0}} \sqrt[n]{a_n} = r$ implies the following given any $\epsilon > 0$.
\begin{align*}
\exists N > 0 \text{ such that } n > N &\implies |\sqrt[n]{a_n} – r| < \epsilon\\
&\implies |\sqrt[n]{a_n}| – |r| < \epsilon \,\, \text{ by the reverse triangle ineq.}\\
&\implies \sqrt[n]{a_n} – r < \epsilon \,\,\text{ since } r,\, a_n \geq 0\\
&\implies r – \sqrt[n]{a_n} > -\epsilon\\
&\implies 1 – \sqrt[n]{a_n} > -\epsilon \,\,\text{ since } r < 1\\
&\implies a_n < (1 + \epsilon)^n\\
&\implies \sum_{n=1}^\infty a_n < \sum_{n=1}^\infty (1 + \epsilon)^n
\end{align*}
But $\displaystyle{\lim_{n\to \infty}} \frac{(1-\epsilon)^{n+1}}{(1 – \epsilon)^n} = 1 + \epsilon > 1$ since $\epsilon > 0$.
So by the ratio test, $\sum_{n=1}^\infty (1 + \epsilon)^n$ diverges and by the comparison test, we can't say anything about
$\sum_{n=1}^\infty a_n$.
So clearly I've hit a snag going down this route. I've been trying to think of alternate routes but I can't think of any. Are there any hints that can be offered?
Best Answer
Choose $\epsilon$ to be small enough so that $r+\epsilon < 1$. Then there exists $N$ such that $|\sqrt[n]{a_n} -r | < \epsilon$ for all $n \ge N$. This implies $\sqrt[n]{a_n} < r+\epsilon < 1$ for all $n \ge N$ so $\sum_{n=N}^\infty a_n \le \sum_{n=N}^\infty (r+\epsilon)^n < \infty$.