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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.

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    $\begingroup$ $|a_n| \le \theta^n$ $\endgroup$ – gammatester Dec 1 '18 at 21:30
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    $\begingroup$ 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 $|a_n|\le \theta$. $\endgroup$ – Bernard Dec 1 '18 at 21:32
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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.

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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...

Always check to make sure you are using the hypotheses of the problem, otherwise things tend to go agley.

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$\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 $\theta \lt1$.

Thus the series converges absolutely by comparison with a geometric series.

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  • $\begingroup$ How do you conclude that it equals $1/1-\theta$ at the end? $\endgroup$ – SacredScout Dec 1 '18 at 23:03
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    $\begingroup$ $(1-\theta)(1+\theta+\theta^2+\dots+\theta ^n)=1-\theta ^{n+1}$. Divide both sides by $(1-\theta) $ and let $n\to\infty$. This is known as "telescoping", because of all the cancellation. $\endgroup$ – Chris Custer Dec 1 '18 at 23:25

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