Let $a_n$ be any sequence converging to $a$ when $n \to \infty$.

Can you rewrite $a_n$ so that it is the sum of two other sequences? $$a_n=b_n + c_n,$$ with $b_n=b$ for every $n \in \mathbb{N}$ and $c_n\to 0$ as $n\to \infty$.

In other words: Is a converging sequence ($a_n$) actually a null sequence ($c_n$) "shifted" by a constant ($b$)?

Or is there any counterexample where one is not allowed to do so?

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    $\begingroup$ You probably mean $b=a$, otherwise $c_n$ cannot converge to $0$. $\endgroup$ – Yves Daoust Jan 2 '16 at 15:57
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    $\begingroup$ "a null sequence" isn't defined. $\endgroup$ – Yves Daoust Jan 2 '16 at 15:57
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    $\begingroup$ @YvesDaoust ProofWiki (proofwiki.org/wiki/Definition:Null_Sequence) would disagree with you. Although I, myself, would have upon hearing "null sequence" out of context would have assumed it meant {a_i} where all a_i = 0. In context, I had no such impression however. $\endgroup$ – fleablood Jan 2 '16 at 16:42
  • $\begingroup$ @fleablood: Ok, my bad. $\endgroup$ – Yves Daoust Jan 2 '16 at 16:43
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    $\begingroup$ Well, I wasn't familiar with the term either. I was going to chastise about how "obviously" a null sequence would mean only zero terms and ... $\endgroup$ – fleablood Jan 2 '16 at 17:07

Set $b_n = a$ and $c_n = a_n - a$ for all $n \in \mathbb{N}$. Then $a_n = b_n + c_n$ for all $n \in \mathbb{N}$ and $c_n = a_n - a \to a-a = 0$ for $n \to \infty$.

  • $\begingroup$ Yeah. It's pretty easy when you can just define c<sub>n</sub> to be the set that satisfies this condition. $\endgroup$ – Daniel Jan 2 '16 at 20:19
  • $\begingroup$ Of course, this only works in spaces where there is actually a subtraction operation. Converging sequences also exist in metric spaces with no such concept! But granted, the question pretty much asserts a vector space structure. $\endgroup$ – leftaroundabout Jan 3 '16 at 17:29
  • $\begingroup$ Any topologcial abelian group should do the trick. (We don’t really need abelian, but then we deal with different sequences, depending on which side we choose for our construction.) $\endgroup$ – Jendrik Stelzner Jan 3 '16 at 17:34

Yes, you can do that. Simply take $b_n=a,c_n=a_n-a$. By basic properties of limits $$\lim\limits_{n\rightarrow\infty}c_n=\lim\limits_{n\rightarrow\infty}(a_n-a)=(\lim\limits_{n\rightarrow\infty}a_n)-a=a-a=0$$


Let $(a_n)$ be a sequence converging towards $a$ and let define the following sequences: $$\forall n\in\mathbb{N},b_n:=a,c_n:=a_n-a.$$ One has: $$\forall n\in\mathbb{N},a_n=b_n+c_n$$ and $(c_n)$ is converging towards $0$.


For a constant $b$ any number $a$ (whether it's a term in a sequence or not) can be written as $a = b + c$ where $c = a - b$ so any sequence $\{a_n\}$ can be written as $\{b + c_n\}$ where $c_n = a_n - b$ and in particular if $\lim a_n = a$ then the sequence can be written as $\{a + c_n\}$ where $c_n = a_n - a$. And clearly $\lim \{a_n\} = \lim \{a + c_n\} = a$.

So your question boils down to does $\lim\{b + c_n\} = b + \lim\{c_n\}$? And therefore if $b = a = \lim a_n$ does $\lim c_n = 0$.

This should be a basic proposition early on in the study of convergent sequence and the answer is: yes.

$|a - a_n| = |(a -b) - (a_n - b)| = |(a - b) - c_n|$. So whatever $\epsilon$, $N$, $n > N$ crap that you can say about $a$ and $a_n$ can also be said about $(a-b)$ and $c_n$.

So if $c_n = a_n - b$ then $\{a_n\} \rightarrow a \iff \{c_n\} \rightarrow (a - b)$.


You are allowed to add, subtract, multiply or divide all terms of a converging sequence with a constant, and you get another converging sequence that converges to $L+C,L-C,L\cdot C, L/C$. $C/L$ also works, provided the original sequence has no zero.

In your case, if $a_n\to a$, then $c_n:=a_n-a\to 0$ and $c_n+a\to a$. You can also have $c_n+b\to a$, provided that $c_n:=a_n-b$, and then $c_n\to a-b$.

  • $\begingroup$ Original sequence must have no zero and the limit has to not be zero. $\endgroup$ – Wojowu Jan 2 '16 at 16:03
  • $\begingroup$ @Wojowu: let's say that the sequence has no zero, even at infinity ;-) $\endgroup$ – Yves Daoust Jan 2 '16 at 16:04

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