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The problem: Suppose $(a_n)_{n\in\mathbb N}$ is a sequence in $\mathbb R$ moreover that $(a_n)_{n\in\mathbb N}$ converges to $L \in \mathbb R$. $(k_n)_{n\in\mathbb N}$ is a sequence in $\mathbb N$, with $\lim_{n \to \infty}\frac{1}{k_n} = 0$. How can we show $(a_{(k_n)})_{n\in\mathbb N}$ converges to L?

My progress: Intuitively I can see why this is true moreover, after seeing a series of classes full of epsilon delta proofs of different limit theorems within sequences in general. I would bet that this will be a proof with creative uses of the epsilon delta definitions perhaps more than once in a nested fashion.

I've spent about 7-8hrs today on this question and have literally made no progress other than trying to bend the definition of convergence around like a mad man.

Florian Suess
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    A note: as @user143234 remarked in his answer, then the convergence of $a_{k_n}$ to $L$ follows from a general theorem in mathematical analysis, which can be found also in stadard textbooks as E. Fisher’s Intermediate real analysis. The general argument follows from the monotonicity of $\limsup$, $\liminf$ and from the fact that since $k_n\to\infty$ then it is possible to extract from it a strictly monotonic subsequence converging to infinity. In sum, this result relies on monotonicity and thus on order arguments. – Daniele Tampieri Aug 02 '18 at 08:56

2 Answers2

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Let $\epsilon >0$. There exists $m$ such that $|a_n-L| <\epsilon$ for $n >m$. There exists $p$ such that $\frac 1 {k_n} < \frac 1 m$ for $n>p$. If $n >p$ then $|a_{k_n} -L| < \epsilon$.

Kavi Rama Murthy
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If $\lim_{n} \frac{1}{k_n} \to 0$, then it is obvious that $\lim_{n} k_n \to \infty$ (you can use simple $\epsilon-\delta$ techniques to prove this). If a sequence converges, then every subsequence converges to the same limit (Prove: If a sequence converges, then every subsequence converges to the same limit.). $\{a_{k_n}\}$ is just a subsequence, so the result follows.

user143234
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    This may be a technicality, but I think it's worth pointing out for these type of exercises: $(k_n){n \geq 1}$ need not be increasing and therefore $a{k_n}$ is not always well defined. – qualcuno Aug 02 '18 at 07:33
  • Yes, thanks for pointing this out. I think strictly speaking, its not a sub-sequence – user143234 Aug 02 '18 at 07:35
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    My answer shows that ${a_{{k_n}}}$ converges to $L$ whether or not it is a subsequence of ${a_n}$. – Kavi Rama Murthy Aug 02 '18 at 07:38
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    The sequence $k_n$ contains an increasing subsequence, so this argument can still be saved. – Dirk Aug 02 '18 at 07:41
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    I think convergence for a sub-sequence does not imply convergence for the general sequence to the same limit – user143234 Aug 02 '18 at 07:47
  • @KaviRamaMurthy In context of your reply that your proof shows ${a_{k_n}}$ converges to $L$ regardless of the fact whether or not it is a subsequence, I would love explanation. In your proof, you say, " $\exists$ some $p$ such that $\frac 1 {k_n} < \frac 1 m$ ". I am interested in knowing how this statement still holds when the sub-index, i.e ${k_n}$ is not convergent? – Subhasis Biswas Aug 02 '18 at 07:59
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    What I am saying that ${k_n}$ need not be strictly increasing to $\infty$. It can even decrease in-between but the hypothesis tells us that $k_n \to \infty$ and this is crucial. In choosing $p$ I have used the fact that $k_n \to \infty$. You can use the term subsequence only when $k_n$ is strictly increasing. – Kavi Rama Murthy Aug 02 '18 at 08:09
  • I get it now! Your proof does allow the sub-index to be partly oscillatory. I messed up on the definition of subsequence. Thank you for your clarification! – Subhasis Biswas Aug 02 '18 at 08:12
  • @user143234 Consider the sequence ${(-1)^n}_n$. Here consider a subsequence with even-index, i.e $2, 4, 6, ...$. We can see that the resulting subsequence is ${1}_n$, which converges to $1$. Now, consider the other subsequence with odd indexing, i.e. ${-1}_n$, which converges to $-1$. We can readily see, that the main sequence doesn't converge anywhere but its subsequences do. Hence, we can safely say, that mere convergence of a single subsequence cannot guarantee the convergence of the parent sequence. However, it is true when every subsequence of a sequence converge to the same limit. – Subhasis Biswas Aug 02 '18 at 08:36