# Show with the direct Comparison Test that $\sqrt[n]{|a_n|}\leq\theta$ converges absolutely

Let $$\sum_{n=1}^{n=\infty}{a_n}$$ be an infinite series of real numbers. There is a $$\theta$$ with $$0<\theta<1$$ and a $$n_0 > 0$$, so $$\sqrt[n]{|a_n|}\leq\theta$$ for all $$n \ge n_0$$. Show that the series converges absolutely.

So I need verification here. Quite funny, because this is the first task out of many and I am certain that the other ones are correct, but this one I am not sure of.

My proof feels wrong, it feels like as if it is way too short.

My proof here:

Since $$\lvert a_{n} \rvert \leq \theta$$ and $$\theta \in (0,1)$$, the series $$\sum_{n=1}^{n=\infty}{a_n}$$ converges. And also since $$\sqrt[n]{|a_n|}≤a_n$$, the series converges absolutely.

• $|a_n| \le \theta^n$ – gammatester Dec 1 '18 at 21:30
• I don't see why $0\le |a_n| \le \theta <1$ implies the series converges. Anyway, the hypothesis is $\sqrt[n]{a_n}\le \theta$, not $|a_n|\le \theta$. – Bernard Dec 1 '18 at 21:32

We have $$\sqrt[n]{|a_n|}\le \theta\to 0\le |a_n|\le \theta^n$$therefore $$0\le \sum_{n=1}^{\infty}|a_n|\le \sum_{n=1}^{\infty}\theta ^n={\theta\over 1-\theta}$$which completes the proof.
We are told that $$|a_n|^{\frac1n}\leq \theta$$ whenever $$n \geq n_0$$. So if $$n \geq n_0$$, it then follows that we have $$0\leq |a_n|\leq\theta^n$$. Since $$\theta \in (0,1)$$, it also follows that $$\sum_{n=1}^\infty \theta^n$$ is a convergent geometric series. Now the comparison test seems relevant...
$$\sqrt[n]{\mid a_n\mid}\lt\theta\implies \mid a_n\mid\le\theta ^n\implies \sum_{n=0}^\infty \mid a_n\mid\le\sum_{n=0}^\infty\theta^n=\frac 1{1-\theta}\lt\infty$$, since $$\theta \lt1$$.
• How do you conclude that it equals $1/1-\theta$ at the end? – SacredScout Dec 1 '18 at 23:03
• $(1-\theta)(1+\theta+\theta^2+\dots+\theta ^n)=1-\theta ^{n+1}$. Divide both sides by $(1-\theta)$ and let $n\to\infty$. This is known as "telescoping", because of all the cancellation. – Chris Custer Dec 1 '18 at 23:25