I'm trying to determine whether $$\sum_{n=1}^\infty \left ( \sqrt[n]{2}-1\right )$$ converges or diverges. Ratio, root, nth term, etc tests are either inconclusive or too difficult to simplify. I feel like there must be something I can bound this series by but I can't think what.
In this question the answerer very smartly (I have no idea how he/ she thought to do that) used the fact that $(1+\frac 1n)^n \le e \le 3$ to bound $\sqrt[n]{3} -1$ by $\frac 1n$ but that only worked because $3\ge e$. So even though that question looks very similar I don't think I can apply that idea here.
Edit: This is in the section before power/ Taylor series so I don't think I'm allowed to use that.
Answer
Hint:
$$\sqrt[n]{2}-1 = e^{\frac{\log{2}}{n}} - 1 = \frac{\log{2}}{n} + O(n^{-2})$$
as $n \to \infty$.
ADDENDUM
To address your specific problem, consider instead
$$\left (1+\frac{\log{2}}{n} \right )^n $$
which you should be able to show is less than $2$. I that case, then you can show that $\sqrt[n]{2}-1 \gt (\log{2})/n$ and draw your conclusion about the sum.
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