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If $x=(x_1.x_2,x_3,...,x_n,...)$ and $\sum^\infty_{k=1} |x_k|^p<\infty$,$p\geq 1$ can you help me to show that $\lim_{k\to\infty}|x_k|^p=0$?

gov
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  • Can you do the case $p=1$? – Gerry Myerson Jul 30 '13 at 11:10
  • I'm afraid I can't :/ – gov Jul 30 '13 at 11:11
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    Sorry, Gorica, if you can't prove that if $\sum a_k$ converges then $a_k\to0$, I don't know why you are even thinking about $\ell^p$ spaces. – Gerry Myerson Jul 30 '13 at 11:19
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    @GerryMyerson Why would the case $p=1$ be any easier? The statement is clearly completely independent of what $p$ is... – not all wrong Jul 30 '13 at 11:22
  • @GerryMyerson If you don't want to help, then don't comment at all. If I forgot some thing because I haven't use them for years, it doesn't mean that I don't understand them and that I couldn't solve them if somebody reminds me how to do that. – gov Jul 30 '13 at 11:24
  • @Sharkos, it's clear to you, and it's clear to me, but I was hoping it would become clear to OP. Unfortunately, OP was in no position to take advantage of my remarks. – Gerry Myerson Jul 30 '13 at 11:26
  • Gorica, I did help, you just weren't ready to appreciate it. – Gerry Myerson Jul 30 '13 at 11:27
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    @GerryMyerson Well, I guess it's subjective whether that's helpful. For someone clearly in a state of confusion already it smacks of misdirection to me. shrug – not all wrong Jul 30 '13 at 11:29
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    @GerryMyerson You didn't. You asked me if I know to show it for $p=1$, I said I don't and that was all of your help. In addition, you wondered why do I EVEN think about $l^p$. You can see down there what "to help" means. – gov Jul 30 '13 at 11:30
  • @Sharkos, I hope everything you ever try to do works out exactly the way you hoped it would. I'm not that lucky. – Gerry Myerson Jul 30 '13 at 12:53
  • Gorica, I still wonder. – Gerry Myerson Jul 30 '13 at 12:54
  • Gerry, keep wondering. And spread your negativity somewhere else. If I achieved a level where I need to think about $l^p$ space it probably means that I passed a level where I had to think about convergent sequences. It's not so bad to forget something or not to know something. It's bad if you don't want to get reminded of that what you forgot or if you don't want to learn something new. And I am pretty sure that there are things that I know and you don't know... Thanks for trying to help me, but you really didn't. – gov Jul 30 '13 at 13:27
  • @GerryMyerson Just like to say: my apologies if I've irritated you! I'm not trying to rub anybody up the wrong way, but everybody's back has been put up here. I've expressed my view that the helpfulness of a comment is subjective and nothing should be taken personally clearly, I hope. I'm also deeply nonplussed by your last comment, but if you mean "It didn't work out as I intended, so don't try to be uppity" I refer you to my apology again! You tried to help, it didn't work, because the OP's mind was in a different place. – not all wrong Jul 30 '13 at 20:06
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    @Gorica I wonder the same thing as Gerry did. I do not think there is anything wrong with that though. To me, it does seem you are not ready for many concepts involving $\ell^p$ spaces. The reasons are 1) the problem could have been phrased without mentioning $p$ -- that is quite a red alert in logical reasoning; and 2) when you leave out $p$, the problem becomes a very simple (and even more general) result from calculus: if $\sum_{n=1}^\infty a_n$ converges, then $\lim_{n\to\infty} a_n = 0$. However, I do feel that Gerry's comments seem a little too harsh. – Tunococ Jul 30 '13 at 21:01
  • @Sharkos, no worries. If you're OK, I'm OK. Gorica, you're welcome. – Gerry Myerson Jul 31 '13 at 00:03

1 Answers1

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As $\sum_{k=1}^\infty|x_k|^p < \infty$ we shall say the series is convergent. Apply Cauchy's criterion.

A sequence $\{x_n\}$ is convergent iff for any $\epsilon > 0$ $\exists$ $m \in \mathbb{N}$ s.t. $|x_p - x_q| < \epsilon$ when $p , q > m$.

Here the sequence of partial sum $\{s_n\}$ is convergent Applying Cauchy's criterion on it we shall get $|s_k -s_l|< \epsilon$ when $p,q > m$.

Take $p=k$ and $l=k-1$

Then you shall get $| |x_k|^p | < \epsilon$ when $k>m$

Which is equivalent to $\lim_{k \rightarrow \infty}|x_k|^p = 0$$

Supriyo
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    If you don't want to use something as "fancy" as Cauchy's criterion (albeit it is definitely a concept you want to know and master), you can also look at a simpler case: set $$S_n\stackrel{\rm{}def}{=} \sum_{k=1}^n |x_k|^p$$ By assumption, $S_n$ converges to some limit $\ell\in\mathbb{R}+$. Then, one can write $$|x_n|^p = S_n - S{n-1}\xrightarrow[n\to\infty]{} \ell - \ell = 0$$ implying that $$|x_n|\xrightarrow[n\to\infty]{} 0$$ as well, i.e. finally $x_n\xrightarrow[n\to\infty]{} 0$. – Clement C. Jul 30 '13 at 11:34