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I intuitively understand proof with limits, but I'm not sure on how to write a formal proof for this example.

For each n $\in$ $\mathbb{N}$, let $a_{n}$ be a real number. Also, let $a_{\infty}$ be a real number. Suppose that $\lim_{n\to \infty} a_{n} = a_{\infty}$.

Prove that if $c \in \mathbb{R}$, then $\lim_{n\to \infty} (a_{n} + c) = a_{\infty} + c$

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  • $\begingroup$ What's your question and where's $b_n$ in the supposed question? $\endgroup$
    – user63181
    Jul 23, 2014 at 19:47
  • $\begingroup$ The last sentence was the question - I edited it so it made sense. $\endgroup$
    – user165987
    Jul 23, 2014 at 19:48
  • $\begingroup$ @user165987 nonetheless, why do you introduce $b_n$ here? $\endgroup$
    – DanZimm
    Jul 23, 2014 at 19:51
  • $\begingroup$ Write down the definition for $\lim_{n\to\infty}a_n=a_\infty$ and then notice that $a_n - a_\infty = a_n+c-c-a_\infty$. $\endgroup$
    – dinosaur
    Jul 23, 2014 at 19:52

1 Answer 1

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Let's take a look at what we're trying to prove: $$ \forall \epsilon > 0 \; \exists N \in \mathbb{N} : \lvert a_n + c - (a_\infty + c) \rvert < \epsilon \; \forall n \ge N \tag{1} $$ Now let's take a look at what we know:

  • Since $a_n \to a_\infty$ we have $$ \forall \epsilon > 0 \; \exists N \in \mathbb{N} : \lvert a_n - a_\infty \rvert < \epsilon \; \forall n \ge N \tag{2} $$
  • $c,a_n \in \mathbb{R} \; \forall n \in \mathbb{N}$

The second statement might seem trivial, but it allows us to use the arithmetic in $\mathbb{R}$ with it.

Now notice that $$ \lvert a_n - a_{\infty} \rvert = \lvert a_n - a_\infty - c + c \rvert = \lvert a_n + c - a_\infty - c \rvert = \lvert a_n + c - (a_\infty + c) \rvert \tag{3} $$ This is the key step to our argument, since this is true we can use the exact same $\epsilon, N$ pair as we used for showing $a_n \to a_\infty$; Now I'll walk you through actually constructing the proof:

Since we're trying to prove something for every $\epsilon > 0$ we must let $\epsilon > 0$ (meaning put $\epsilon$ to be some arbitrary real number so that it is $> 0$). From here we need to show the existence of some $N \in \mathbb{N}$ that satisfies $(1)$. Note that from $(3)$ we have that $\lvert a_n + c - (a_\infty + c) \rvert$ is just as small as $\lvert a_n - a_{\infty} \rvert$ for every $n$. In particular by $(2)$ we know that there exists an $N \in \mathbb{N}$ so that $\lvert a_n - a_{\infty} \rvert < \epsilon \; \forall n \ge N$. Combining these last two facts we get that there exists $N \in \mathbb{N}$ so that $$ \lvert a_n + c - ( a_{\infty} + c ) \rvert = \lvert a_n - a_{\infty} \rvert < \epsilon \; \forall n \ge N \\ \implies \lvert a_n + c - ( a_{\infty} + c ) \rvert < \epsilon \; \forall n \ge N $$ which finally shows that $a_n + c \to a_\infty + c$.


More concisely this can be put as:

Let $\epsilon > 0$. Since $a_n \to a_\infty$ we know $\exists N \in \mathbb{N}$ so that $$ \lvert a_n - a_{\infty} \rvert < \epsilon \; \forall n \ge N, \lvert a_n + c - ( a_{\infty} + c ) \rvert = \lvert a_n - a_{\infty} \rvert \\ \implies \lvert a_n + c - ( a_{\infty} + c ) \rvert < \epsilon \; \forall n \ge N $$

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