1. [SOLVED] Infinite series convergence

Does $\sum^\infty _{n=1} \sqrt[n]{2}-1$ converge?
I am sure that $\sum^\infty _{n=1} \sqrt[n]{2}$ diverges, since
$\lim \limits_{n \to \infty} \sqrt[n]{2}=1$, but $\lim \limits_{n \to \infty} \sqrt[n]{2}-1=0$....

2. $\sqrt[n]{2}-1\to0$ and $\frac{\ln2}n\to0$ as $n\to\infty,$ hence $\frac{\ln 2}{n}>\frac{1}{n}$ and your series diverges.

3. Originally Posted by Krizalid
$\sqrt[n]{2}-1\to0$ and $\frac{\ln2}n\to0$ as $n\to\infty,$ hence $\frac{\ln 2}{n}>\frac{1}{n}$ and your series diverges.

I am confused, how does $\frac{\ln 2}{n}>\frac{1}{n}$ imply that $\sum^\infty _{n=1} (\sqrt[n]{2}-1)$diverges? by comparison?

4. Originally Posted by akolman
Does $\sum^\infty _{n=1} \sqrt[n]{2}-1$ converge?
I am sure that $\sum^\infty _{n=1} \sqrt[n]{2}$ diverges, since
$\lim \limits_{n \to \infty} \sqrt[n]{2}=1$, but $\lim \limits_{n \to \infty} \sqrt[n]{2}-1=0$....
Note that

\begin{aligned}\forall{n}\in\mathbb{R}~\sqrt[n]{2}&=2^{\frac{1}{n}}\\
&=e^{\frac{\ln(2)}{n}}\\
&=1+\frac{\ln(2)}{n}+\cdots\end{aligned}

So

\begin{aligned}\forall{n}\in\mathbb{R}~\sqrt[n]{2}-1&=1+\frac{\ln(2)}{n}+\cdots-1\\
&=\frac{\ln(2)}{n}+\cdots\geqslant\frac{\ln(2)}{n} \end{aligned}

So

$\sum_{n=1}^{\infty}\frac{1}{n}\to\infty\leqslant\s um_{n=1}^{\infty}\frac{\ln(2)}{n}$

So your series is greater than a divergent series, thus divergent.