How to show that the series $$ \sum_{n=1}^\infty (\sqrt[n]{2}-1)$$ diverges ?
Asked
Active
Viewed 1.9k times
9
-
Any n-th root of 2 will always be greater than 1. – hjpotter92 Mar 19 '13 at 07:41
-
This old question and this other old question are basically duplicates. – JimmyK4542 Sep 18 '17 at 18:37
3 Answers
11
Hint: $$2^{1/n}-1=e^{\frac{1}{n}\log 2}-1\sim \frac{1}{n}\log 2.$$
-
3
-
1@ElmarZander Your inequality is usefull in our case of divergence, but by equivalence relation we can prove both divergence and convergence in different situation. – Mar 19 '13 at 08:51
10
Observe that $$\left(1+\frac{1}{2n}\right)^n=\sum\limits_{i=0}^n\binom{n}{i}\frac{1}{2^in^i}$$ and that for $i\ge 1$ we have $\binom{n}{i}\le n^i$ so this becomes $$\left(1+\frac{1}{2n}\right)^n\le 1+\sum\limits_{i=1}^n\frac{1}{2^{i}}<2$$ and thus we have $$\sqrt[n]{2}>1+\frac{1}{2n}$$ which gives us $$\sum\limits_{n=1}^\infty (\sqrt[n]{2}-1)\ge\sum\limits_{n=1}^\infty\frac{1}{2n}=\infty.$$
Alex Becker
- 60,569
2
$\lim_{x \to 0}\frac{2^x-1}{x} = \log2$. This implies $\lim_{n \to +\infty} \frac{2^\frac{1}{n}-1}{\frac{1}{n}} = \log2$. Since the series $\sum_{n=1}^{+\infty} \frac{1}{n}$ is divergent, so by Comparison test (limit form), the series $\sum_{n=1}^{+\infty} (\sqrt[n]{2}-1)$ is divergent.
