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Let $(a_k)_{k \in \mathbb{N}}$ be some real sequence. Show that if $\lim_{k \rightarrow \infty} \sup (\sqrt[k]{|a_k|}) < 1$ then $\sum_{k=0}^\infty a_k$ converges absolutely.

I have some general questions about the limes supremum first.

$\lim_{k \rightarrow \infty} \sup (\sqrt[k]{|a_k|})$ is defined as $\lim_{k \rightarrow \infty} (\sup(\bigcup_{i=k}^\infty (\sqrt[i]{|a_i|}))$

How exactly is the union of the set to be interpreted? $\bigcup_{i=k}^\infty (\sqrt[i]{|a_i|})$ is the union of all $\sqrt[i]{|a_i|}$ for all i larger than some k, right?

Now the supremum of that union is the largest term of that set. So I think this supremum depends on the choice of k. Is the supremum of the union therfore loosely speaking a function of k? And if I let k go to infinity I get the limes supremum, correct?

Now in the proof I don't understand how that limit could be less than 1. Since $\forall c \gt 0 \in \mathbb{R}: \lim_{k=0}^\infty \sqrt[k]{c_k} = 1$ So how can $\lim_{k=0}^\infty \sqrt[k]{|a_k|} < 1$?

And I guess the proof somehow has to make use of the root test. But if I take $\sup(\bigcup_{i=k}^\infty (\sqrt[i]{|a_i|})$, which yields a term of the form $\sqrt[k_0]{|a_{k_0}|})$ and then I let $k_0 \rightarrow \infty$, and by assumption this limit is less than 1, does it immidiately follow by the root test that $\sum_{k=0}^\infty a_k$ converges absolutely?

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  • $\begingroup$ It's not the limes superior of sets here, but the limes superior of a sequence, $$\limsup_{k\to\infty} \sqrt[k]{\lvert a_k\rvert} = \lim_{k\to\infty} \left(\sup \{ \sqrt[n]{\lvert a_n\rvert} : n \geqslant k\}\right).$$ $\endgroup$ – Daniel Fischer Jun 15 '14 at 13:29
  • $\begingroup$ What I don't understand is that you take the supremum of the set $\{\sqrt[n]{|a_n|}: n \geq k \}$ for a given k. And then you take the limit of k? I don't see how that makes sense. The k is given and you take the supremum, then how can you take the limit of k after that??? $\endgroup$ – eager2learn Jun 15 '14 at 13:35
  • $\begingroup$ For each $k$, we have a set $S_k = \{ \sqrt[n]{\lvert a_n\rvert} : n \geqslant k\}$. Then $b_k = \sup S_k$ is a number. And since $S_{k+1} \subset S_k$, it follows that $b_{k+1} \leqslant b_k$. Then $\limsup\limits_{k\to\infty} \sqrt[k]{\lvert a_k\rvert} = \lim\limits_{k\to\infty} b_k$. $\endgroup$ – Daniel Fischer Jun 15 '14 at 13:42
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Without loss of generality, we may assume that $a_{n}\geq0$ (otherwise, replace $a_{n}$ by $|a_{n}|$). Suppose that $\limsup_{n}\sqrt[n]{a_{n}}=l<1$. For each $n$, let $b_{n}=\sup\left\{ \sqrt[k]{a_{k}}\mid k\geq n\right\} $. Choose $r$ such that $l<r<1$. Note that $\lim_{n}b_{n}=l<r$, so there exists $N$ such that $b_{n}<r$ whenever $n\geq N$. In particular, for any $n\geq N$, $\sqrt[n]{a_{n}}\leq b_{n}<r$. It follows that $a_{n}\leq r^{n}$. Note that $\sum_{n=N}^{\infty}r^{n}$ is a geometric series, which converges because $|r|<1$. By comparison test, $\sum_{n=N}^{\infty}a_{n}$ also converges.

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