(Proof Verification) If $\lim \sup \sqrt[k]{x_k}>1$, then $\sum_{k=1}^{\infty}x_k = \infty$.

Question

Let $x_k \geq 0 $ for all $k\in\mathbb{N}$. Note that $\lim_{k\to\infty}\sqrt[k]{x_k}$ need not exist.

If $\lim \sup \sqrt[k]{x_k}>1$, then $\sum_{k=1}^{\infty}x_k = \infty$.

Here is my attempt on proving this:

$\lim \sup \sqrt[k]{x_k}>1$ implies that for any $n$th tail of the sequence $\sqrt[k]{x_k}$, namely $s_n$, $\sup\{s_n\} > 1$ for any $n\in \mathbb{N}$. Then this implies that there will always be $\sqrt[k]{x_k} > 1$ for some $k \in\mathbb{N}$ and the sequence $\sqrt[k]{x_k} \not\to 0 $ and consequently $x_k \not\to 0$ as $k \to\infty$. Therefore, it fails the $n$th term test and the series $\sum_{k=1}^{\infty}x_k = \infty$. $\blacksquare$

Answer 1

You write "$\sqrt[k]{x_k} \not\to 0$ and consequently $x_k \not\to 0$ as $k \to \infty$" - this is false, as seen for the sequence $(\frac 1 k)_{k \ge 1}$.

One possible correct answer is the following: if $\limsup \sqrt[k] {x_k} = l > 1$ this means that there exist a subsequence of $(x_{k_n}) _{n \ge 1}$ such that $\sqrt[k_n] {x_{k_n}} \to l$. Let $y_n = x_{k_n}$. Notice that $\sum x_k \ge \sum x_{k_n} = \sum y_n$ because you remove all the terms that are not of the form $x_{k_n}$ from the original series (which has positive terms). By the root test, since $\sqrt[n] {y_n} \to l$, it follows that $\sum y_n$ diverges, therefore $\sum x_n$ diverges, too.

Notice that the proof does not explicitly use the definition of $\limsup$; the only important thing is that $(\sqrt[k] {x_k})_{k \ge 1}$ has a limit point strictly larger than $1$ (which, of course, is equivalent to $\limsup \sqrt[k] {x_k} > 1$).

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An alternative solution is to notice that the subsequence $(x_{k_n})_{n \ge 1}$ considered above has the property $\sqrt[k_n] {x_{k_n}} > 1$ from a certain $n_0$ onwards, which implies $x_{k_n} > 1$ for $n \ge n_0$, which in turn implies $x_{k_n} \not\to 0$, therefore $x_n \not\to 0$, so by the zero test $\sum x_n$ diverges.