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"Let $(a_n)_{n=m}^{\infty}$ be a sequence of real numbers, let $c$ be a real number, and let $k \geq 0$ be a non-negative integer. Show that $(a_n)_{n=m}^{\infty}$ converges to $c$ iff $(a_{n+k})_{n=m}^{\infty}$ converges to $c$."

My attempt to prove this:

$\Rightarrow$ Let $\varepsilon >0$ be an arbitrary positive real number. Then, since $(a_n)_{n=m}^{\infty}$ converges to $c$, by definition of convergence we have that there exists an integer $N \geq m$ s.t. $|a_n-c|\leq \varepsilon$ $\forall n \geq N$. Now, if we suppose for the sake of contradiction that $(a_{n+k})_{n=m}^{\infty}$ doesn't converge to $c$ this means there exists an integer $N'\geq N$ s.t. $|a_{N'+k}-c|>\varepsilon$ (if this wasn't the case, $(a_{n+k})_{n=m}^{\infty}$ would be convergent by definition since we would have $|a_{n+k}-c|\leq \varepsilon$ $\forall n \geq N$), but, taking $p:=N'+k\geq N$ this implies $|a_p-c|> \varepsilon$, a contradiction. So, if $(a_n)_{n=m}^{\infty}$ converges to $c$, then $(a_{n+k})_{n=m}^{\infty}$ converges to $c$ too.

Now, I'm having a hard time proving the leftward implication ($(a_{n+k})_{n=m}^{\infty}$ converges to $c$ implies $(a_n)_{n=m}^{\infty}$ converges to $c$ too) so I would appreciate any hint about how to carry out that part of the proof (and also any comment about my proof so far).

EDIT: I'm now trying to prove the leftward implication by showing that $(a_{n+k})_{n=m}^\infty = (a_n)_{n=m+k}^\infty$ and then using the fact (which I already proved) that $(a_n)_{n=m}^\infty \to $ $c$ as $n \to \infty$ iff $(a_n)_{n=m'\geq m}^\infty \to $ $c$ as $n \to \infty$ but here too I'm having some difficulties in proving the equality (I tried induction on $k$ but I haven't been able to carry out the inductive step; trying to argue by contradiction didn't work either.)

Any help would be appreciated.

Best regards,

lorenzo.

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  • $\begingroup$ Hint: take $N'=N+k$. $\endgroup$ – Yves Daoust Oct 4 '15 at 13:32
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Actually, the first version of the question contained perfectly valid (if somewhat verbose) proof of the following theorem.

Theorem 1. The sequence of real numbers $(a_n)_{n=m}^\infty$ converges to $c \in \mathbb R$ if and only if the sequence $(a_n)_{n=m'}^\infty$ for $m' \geq m$ also converges to $c$.

If you observe that the definition of convergence you are using in your proof

Definition 1. The real sequence $(a_n)_{n=m}^\infty$ converges to $c \in \mathbb{R}$ if and only if for every $\varepsilon>0$ there exists $N \in \mathbb N$, $N>m$ such that $$n \geq N \implies |a_n-c| \leq \varepsilon.$$

can be formulated alternatively as

Definition 2. The real sequence $(a_n)_{n=m}^\infty$ converges to $c \in \mathbb R$ if and only if for every $\varepsilon>0$ it contains subsequence $(a_n)_{n=m'}^\infty$ where $m'>m$ that lies completely within the set $B_{c,\varepsilon}=\{z \in \mathbb R \mid |z-c| \leq \varepsilon\}$.

than the proof reduces to few sentences:

  • In one direction, if $(a_n)_{n=m'}^\infty$ converges to $c$, the subsequence $(a_n)_{n=m''}^\infty$, $m''>m'$ of this sequence from Definition 2 that lies within $B_{c,\varepsilon}$ is also a subsequence of $(a_n)_{n=m}^\infty$.

  • In the other direction, if $(a_n)_{n=m}^\infty$ converges to $c$, the subsequence $(a_n)_{n=m''}^\infty$, $m''>m$ of this sequence from Definition 2 that lies within $B_{c,\varepsilon}$ might not be a subsequence of $(a_n)_{n=m'}^\infty$, when $m''<m'$. But, even then, the subsequence $(a_n)_{n=max\{m',m''\}}^\infty$ certainly is.


To see that it really makes no difference if we replace the $(a_n)_{n=m'}^\infty, m'>m$ with $(a_{n+k})_{n=m}^\infty, k>0$ in the theorem 1, consider the following proof of the amended theorem

  • In one direction, if $(a_{n+k})_{n=m}^\infty$ converges to $c$, the subsequence $(a_{n+k})_{n=m''}^\infty$, $m''>m'$ of this sequence from Definition 2 that lies within $B_{c,\varepsilon}$ is also a subsequence of $(a_n)_{n=m}^\infty$.

  • In the other direction, if $(a_n)_{n=m}^\infty$ converges to $c$, the subsequence $(a_n)_{n=m''}^\infty$, $m''>m$ of this sequence from Definition 2 that lies within $B_{c,\varepsilon}$ might not be a subsequence of $(a_{n+k})_{n=m}^\infty$, when $m''-m<k'$. But, even then, the subsequence $(a_n)_{n=max\{m+k,m''\}}^\infty$ certainly is.

Note that both sequences $(a_n)_{n=m+k}^\infty$ and $(a_{n+k})_{n=m}^\infty$ are obtained by removing first $k$ elements from the sequence $(a_n)_{n=m}^\infty$.

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  • $\begingroup$ that's not the theorem for which I'm asking help; it is $(a_{n+k})_{n=m} ^\infty \to $ $c$ as $n$ $ \to \infty \Leftrightarrow (a_n)_{n=m} ^\infty \to$ $ c$ as $n$ $\to \infty$ ($k \geq 0$ non-negative integer). It's my fault, the text of the answer contained an error, sorry for that. Check now. $\endgroup$ – lorenzo Oct 6 '15 at 17:12
  • $\begingroup$ It doesn't really make any difference. The sequences $(a_{n+k})_{n=m}^\infty$ and $(a_n)_{n=m+k}^\infty$ are essentially one and the same sequence. The only difference is that one is indexed by the set $\{m,m+1,\ldots\}$ and the other by the set $\{m+k,m+k+1,\ldots\}$. $\endgroup$ – Zoran Loncarevic Oct 6 '15 at 17:20
  • $\begingroup$ That $(a_{n+k})_{n=m}^ \infty = (a_n)_{n=m+k}^ \infty$ is intuitively clear; I'm having difficulties in proving that equality rigorously. $\endgroup$ – lorenzo Oct 6 '15 at 17:29
  • $\begingroup$ You can replace $(a_n)_{n=m'}^\infty$ with $(a_{n+k})_{n=m}^\infty$ and $m'$ with $m+k$ in the proof above, and it would still hold. $\endgroup$ – Zoran Loncarevic Oct 6 '15 at 17:35
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$(a_{n+k})_{n=m}^\infty\to L\in\mathbb{R}\ ($as $n\to\infty)\overset{DEF.}{\Rightarrow} $ if we take $\varepsilon>0$ arbitrary then $\exists N\geq m$ such that $|a_{n+k}-L|\leq\varepsilon\ \forall n\in\mathbb{N}, n\geq N\Rightarrow |a_{n+k}-L|\leq\varepsilon\ \forall n\in\mathbb{N}, n\geq N-k\Rightarrow |a_{n+k}-L|\leq\varepsilon\ \forall n\in\mathbb{N}, n+k\geq N$ so if we take $j:=n+k$ we have $|a_j-L|\leq\varepsilon\ \forall j\in\mathbb{N}, j\geq N$. Since $\varepsilon>0$ was chosen arbitrarily we can conclude that $\forall \varepsilon >0\ \exists N\geq m$ such that $|a_j-L|\leq\varepsilon\ \forall j\in\mathbb{N}, j\geq N$ so, by definition, $(a_j)_{j=m}^\infty \to L$ (as $j\to\infty$).

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