# Thread: [SOLVED] Infinite series convergence

1. ## [SOLVED] Infinite series convergence

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

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

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

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

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

\displaystyle \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

\displaystyle \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

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

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