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Prove the $n$th Root Test: Suppose $x_k \geq 0$ for all $k\in\mathbb{N}$, and suppose $\sqrt[k]{x_k}\to L$ as $k\to\infty$. Then we have the following conclusions.

I only proved the first part, which states if $L>1$, then the series $\sum_{k=1}^{\infty}x_k$ must diverge. Here is how I did it:

Suppose $\sqrt[k]{x_k}\to L$ where $L>1$. Then, there exists $N\in\mathbb{N}$ such that for all $k \geq N$, $\sqrt[k]{x_k}>r$ where $1 r^N \\ x_{N+1} > &> r^{N+1} \\ &\vdots \\ x_{N+k} &> r^{N+k} \end{align*}

which implies $\sum_{k=N}^{\infty}x_k > \sum_{k=N}^{\infty}r^k$. Since $\sum_{k=N}^{\infty}r^k$ is a geometric series with $r>1$, it must be divergent. Then by the comparison test, $\sum_{k=N}^{\infty}x_k$ must be also divergent. Thus, $\sum_{k=1}^{\infty}x_k = > \sum_{k=1}^{N-1}x_k+\sum_{k=N}^{\infty}x_k$ is divergent.

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    "Then by the comparison test"... comparison of what to what? you only need the fact that once you know $\sqrt[k]{x_k}>r>1$, $x_k\ge \sqrt[k]{x_k}>r>1$ to conclude – 2017-02-10
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    I am comparing $x_k$ to $r^k$. Since $x_k>r^k$ for $k \geq N$ for some $N\in\mathbb{N}$, doesn't that mean the series of $r^k$ being divergent imply the series of $x_k$ must be also divergent? – 2017-02-10
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    But you wrote that $\sum_{k=N}^{\infty} \sqrt[k]{x_k}$ must be divergent – 2017-02-10
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    Okay, either way I think that was just a typo, you got the right idea. But from your wording I thought you were saying something along the line of $\sum x_k$ is divergent, so by comparison test $\sum \sqrt[k]{x_k}$ is divergent, which makes no sense in many ways. – 2017-02-10
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    Yes, I realized it was a typo. I am not trying to prove $\sum_{k=1}^{\infty}\sqrt[k]{x_k}$ is divergent, I am trying to prove $\sum_{k=1}^{\infty}x_k$ is divergent. – 2017-02-10
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    Just to be clear, with your apparent typo fixed, your proof is correct. For the other side, an almost identical proof works. – 2017-02-10

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