Given $\sum_{n=1}^{\infty} a_{n}$ let $\lim sup (a_{n})^{1/n}=\alpha \in \mathbb{R}$.
a) If $\alpha<1$, then $\sum_{n=1}^{\infty} a_{n}$ converges.
b) If $\alpha>1$ then $\sum_{n=1}^{\infty} a_{n}$ diverges.
Definition:
$\lim sup=\lim_{n \to \infty} sup_{k \geq n} a_{k}$
I have already been able to proof a) with this definition.
Proof:
Let $\beta-\alpha>0$, then by definition $\exists N$ s.t. $n \geq N \implies \lvert sup_{k \geq n} (a_{k})^{1/k}-\alpha \rvert<\beta-\alpha$, so $-(\beta-\alpha)<sup_{k \geq n} (a_{k})^{1/k}-\alpha<\beta-\alpha \implies sup_{k \geq n} (a_{k})^{1/k}<\beta$.
Now let $n=N$, then $(a_{k})^{1/k} \leq sup_{k \geq N} (a_{k})^{1/k}<\beta$ for all $k \geq N$. But then $a_{k}<\beta^{k}$, and so $\sum_{n=1}^{\infty} a_{n}$ converges by the comparison with the geometric sequence.
Now I've tried a similar thing for b).
Proof attempt:
Let $\alpha-\beta>0$, then by definition $\exists N$ s.t. $n \geq N \implies \lvert sup_{k \geq n} (a_{k})^{1/k}-\alpha \rvert<\alpha-\beta$, so $-(\alpha-\beta)<sup_{k \geq n} (a_{k})^{1/k}-\alpha<\alpha-\beta \implies \beta<sup_{k \geq n} (a_{k})^{1/k}$.
The problem is now that I only know that this supremum is larger than $\beta$, but I need to show that the series can be bounded below by the divergent geometric series.
Am I on the right track here? How can I show this given my definition of $\lim sup$?
Thanks very much for any help.
I will say you are on the right track (except using $\beta$. You can use $1+\beta$ for $\beta$ small enough where you wrote $\beta$). Next thing you can do is, think about the definition of the supremum, and find a subsequent of $(a_n)_n$ that goes infinity. It is as follows :
If $\alpha>1$, then for any given $0<\epsilon<\alpha-1$, there exists $N\in\mathbb{N}$ such that for any $n\geq N$, $\sup_{k\geq n}(a_k)^{1/k}>1+\epsilon$, i.e. $\exists k_n\geq n$ such that $(a_{k_n})^{1/{k_n}}>1+\epsilon$ for each $n\geq N$. Therefore $a_{k_n}>(1+\epsilon)^{k_n}$, and by letting $n\rightarrow \infty$ you get $a_{k_n}\rightarrow \infty$ (because $k_n\rightarrow \infty$ since $k_n\geq n$). Since $(a_{k_n})_{n\geq N}\subset(a_n)_{n\in\mathbb{N}}$, $(a_n)_{n\in\mathbb{N}}$ does not converge, i.e. diverge.
edit : I will try to write more precisely. Take any $\epsilon$ such that $0<\epsilon<\alpha-1$. Then $$\exists N\in\mathbb{N},\forall n\in{\mathbb{N}}(\sup_{k\geq n}(a_k)^{1/k}>1+\epsilon).$$ Then $$\forall n\in\mathbb{N}(n\geq N\rightarrow\exists k_n\in{\mathbb{N}}(k_n\geq n\land a_{k_n}>(1+\epsilon)^{k_n}))$$ since for $\textbf{each}$ $n\geq N$, $\sup_{k\geq n}(a_k)^{1/k}>1+\epsilon$. Then now we have $(a_{k_n})_n$ that goes infinity by $n\rightarrow \infty$.