Prove that $\sum_{n\geq 1}(\sqrt[n]{n}-1)$ diverges.

calculusreal-analysissequences-and-series

I am studying for an admission exam and I find this problems together. I previously showed that $$\sum_{n\geq 1}(\sqrt[n]{n}-1)^n$$ converges using root test and showing that $\lim_{n\to\infty}\sqrt[n]{n}=1$.

The next exercise is prove the divergence of $$\sum_{n\geq 1}(\sqrt[n]{n}-1)$$ I don`t know if the first result is useful for this.

First, i tried to find a sequence $a_n$ such that $a_n<\sqrt[n]{n}-1$ and $\sum_{n\geq 1}a_n$ diverges, but I can´t find any. Then I tried to use the partial sums: $$\sum_{n=1}^{k}(\sqrt[n]{n}-1)=(1-1)+(\sqrt{2}-1)+\dots +(\sqrt[k]{k}-1)=1+\sqrt{2}+\sqrt[3]{3}+\dots +\sqrt[k]{k}-k$$ but I didn`t know how to continue. My last attempt was try to prove that the partial series are estrictly increasing and unbounded, but I can´t prove that is unbounded. I am stuck at this point. Thanks for the help, and sorry about my english.

Best Answer

I will use a comparison to show that $\sum_{n=1}^{\infty} (\sqrt[n]{n} - 1)$ diverges.

To start, it is well-known that the infinite series $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges. This is the sequence we will be comparing to.

First, we need to prove that $\frac{1}{x}$ is less than $x^{1/x}-1$ at some $x=k$, and that it stays less. So, we want to find when:

$\frac{1}{x} < x^{1/x}-1$

$1 + \frac{1}{x} < x^{1/x}$

$(1 + \frac{1}{x})^{x} < x$ (both sides are positive as we are only considering $x>0$)

This is obviously true for some $x=k$; it is well-known that $(1+\frac{1}{x})^{x}$ is monotonically increasing and approaches $e$ as $x \to \infty$, and $x$ trivially approaches $\infty$ as $x \to \infty$. We can take any $x>e$, say $3$, and this is a point where the terms of $\sum_{n=1}^{\infty} n^{1/n}-1$ are greater than the terms of $\sum_{n=1}^{\infty} \frac{1}{n}$. They will stay greater as $n \to \infty$, and hence we have that:

$\sum_{n=3}^{k} > \sum_{n=3}^{k} \frac{1}{n}$

and as $k \to \infty$, because the right-hand sum diverges, the left-hand sum will diverge. Adding in the $n=1$ and $n=2$ terms will not impact divergence, and the proof is done.

Sidenote: I'm sorry if my proof was badly worded or not rigorous enough, I am fresh out of a calc 2 class and I do not have that much experience writing formal proofs. Hopefully my meaning is clear. Please suggest any edits if it is not.

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