Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to prove: $\{c_n\}$ converges to $c$ if and only if $\{c_n -c\}$ converges to $0$

so I start with ( $\{c_n\}$ converges to $c$ ) $\Rightarrow$ ( $\{c_n -c\}$ converges to $0$ ) and I assume $\{c_n\}$ converges to $c$.

So then $$\lim_{n\to\infty}\{c_n -c\} =\lim_{n\to\infty}\{c_n\} - \lim_{n\to\infty}\{c\} $$ since $$\lim_{n\to\infty}\{c_n\} = c $$ by assumption and $c$ is a constant so $\lim_{n\to\infty}\{c\} = c$ the limit of $$\lim_{n\to\infty}\{c_n -c\} = c- c = 0$$

next I go to ( $\{c_n -c\}$ converges to $0$) $\Rightarrow$ ( $\{c_n\}$ converges to $c$) and I assume $\{c_n -c\}$ converges to $0$

so $$\lim_{n\to\infty}\{c_n -c\} =\lim_{n\to\infty}\{c_n\} - \lim_{n\to\infty}\{c\} = 0$$ and again $c$ is a constant so $$\lim_{n\to\infty}\{c\} = c$$

thus $$\lim_{n\to\infty}\{c_n -c\} =\lim_{n\to\infty}\{c_n\} - c = 0$$ add $c$ to both sides and I get $$\lim_{n\to\infty}\{c_n\} = c$$

I think this is basically correct. I show both directions of the if and only if. Can anybody see any mistakes in my reasoning? I have a Theorem from the textbook for the separation of the limit into two limits and I'm allowed to say the limit of a constant is that constant directly. I know nothing else about $\{c_n\}$ and $c$ is an arbitrary number in $\mathbb{R}$.

share|cite|improve this question
Looks good to me. – Alex Becker Jan 18 '12 at 1:45
+1 for lots of effort! – JavaMan Jan 18 '12 at 2:34
up vote 3 down vote accepted

It's basically correct, but I think it can be improved. It seems that you are assuming in the second part that $\lim\limits_{n\rightarrow\infty} c_n$ exists at the outset. But you can't do this.

What you do is start with two limits that you know exist. Then you can say the limit of the sum exists and is the sum of the limits.

More formally:

If $\lim\limits_{n\rightarrow\infty} c_n=L$ and if $\lim\limits_{n\rightarrow\infty} d_n=M$, then $\lim\limits_{n\rightarrow\infty} (c_n+d_n)$ exists and is equal to $M+L$: $$ \lim\limits_{n\rightarrow\infty}(c_n+d_n) = \lim\limits_{n\rightarrow\infty} c_n+\lim\limits_{n\rightarrow\infty} d_n=L+M. $$

So I would do the second part as follows:

We know that $\lim\limits_{n\rightarrow\infty}(c_n-c)=0$ and that $\lim\limits_{n\rightarrow\infty} c = c$.

Then, $\lim\limits_{n\rightarrow\infty}\bigl ( (c_n-c)+c\bigr) $ exists. But, $(c_n-c)+c=c_n$; so $\lim\limits_{n\rightarrow\infty} c_n$ exists and $$ \lim\limits_{n\rightarrow\infty} c_n = \lim\limits_{n\rightarrow\infty} \bigl ((c_n-c)+c\bigr)= \lim\limits_{n\rightarrow\infty} (c_n-c) +\lim\limits_{n\rightarrow\infty} c =0+c=c. $$

share|cite|improve this answer
thanks, it makes sense that I can't assume the two limits exists when I separate the combined limit. I must have misread that part of the theorem. – AvatarOfChronos Jan 18 '12 at 3:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.