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How would I solve:

$\lim \limits_{n \to \infty} 1+(-1)^n$

Can I split $1$ and $(-1)^n$ up into two different limits added together?

$\lim \limits_{n \to \infty} 1+(-1)^n$ = $\lim \limits_{n \to \infty} 1$ + $\lim \limits_{n \to \infty} (-1)^n$

And if I did this, how would I solve for $\lim \limits_{n \to \infty} (-1)^n$

Thank you

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  • $\begingroup$ What id $1+(-1)^n$ if $n$ is even? If $n$ is odd? Does th elimit exist? $\endgroup$ Mar 29 '15 at 19:45
  • $\begingroup$ This limit does not exist, you must find an $\epsilon>0$ so that there does not exist any $N$ with the required property. $\endgroup$ Mar 29 '15 at 19:45
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Note $(-1)^n$ is $1$ for even $n$ and $-1$ for odd $n$. So $1 + (-1)^n$ is $2$ for even $n$ and $1$ for odd $n$. Therefore, if $a_n = 1 + (-1)^n$, then $a_{2n} = 2$ and $a_{2n+1} = 0$. Since the subsequences $a_{2n}$ and $a_{2n+1}$ have different limits ($\lim a_{2n} = 2$ and $\lim a_{2n+1} = 0$), the limit $\lim a_n$ does not exist.

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The limit does not exist and that can be proven in more than one way.

Noting, as @kobe states, we can find two convergent subsequences to different limits and thus we have a non-convergent sequence at hand. This can be states in terms of the limit superior and limit inferior and noting that they are different.

Another way to prove this is using contradiction. Since the sequence only takes the two values $0$ and $2$, one can trivially see (and prove) that any value except these two cannot be the limit. Assuming $l=0$, and picking up $\varepsilon<2$, we see that the limit cannot exist since $|x_n-x_{n+1}|=2>\varepsilon$. We prove it similarly for $l=2$.

QED.

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Alas, this limit does not exist.

The idea behind a limit is that as you change some parameter (in this case making $n$ larger), the expression becomes a better and better approximation of some value. We call that value the limit.

In this case, the value just oscillates. $1+(-1)^{2n} = 2$ and $1+(-1)^{2n-1} = 0$. So the sequence as $n$ grows looks like:

$$ 0, 2, 0, 2, 0, 2, \cdots $$

That doesn't make closer and closer approximations of anything. Ergo the limit doesn't exist.

Somewhat more formally, we say that if we let $\varepsilon = 1$, then there's no number $L \in \mathbb{R}$ for which there's an $N \in \mathbb{N}$ such that all $n > N$ yields $\left|(1+(-1)^{n}) - L\right| < \varepsilon$. But this is just a formal restatement of what I said above.

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If $\lim_{n \to \infty} 1$ and $\lim_{n \to \infty} (-1)^n$ both existed, you could break up your desired limit as you show. However, the second limit does not exist, so your method does not work.

To find the limits (if any), just consider the sequences. $\lim \limits_{n \to \infty} (-1)^n$ comes from the sequence $-1,1,-1,1,-1,1,\ldots$. This clearly never "settles down" to a single number, so the limit does not exist.

Your desired limit $\lim \limits_{n \to \infty} [1+(-1)^n]$ comes from the sequence $0,2,0,2,0,2,\ldots$. Again, the sequence does not settle down, and there is no limit.

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$$(-1)^n=e^{i\pi n}=\cos(n\pi)+i\sin(n\pi)$$

Does $$\lim_{n\to\infty} \cos(n\pi)\text{ or }\lim_{n\to\infty}i\sin(n\pi)$$ exist?

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You can also think of it like this. We don't know if $\lim_{n \to ∞} (-1)^n$ exits. So we'll set y = $\lim_{n \to ∞} (-1)^n$ and see what happens.

$y = \lim_{n \to ∞} (-1)^n$ (let)

or, $\ln y = \lim_{n \to ∞} n[\ln (-1)]$

$\Rightarrow$ RHS does not exist [∵ $\ln (x)$ is only defined when $x = (0, \infty)$]

$\Rightarrow \lim_{n \to ∞} (-1)^n $ does not exist.

$\Rightarrow \lim_{n \to ∞} 1 + (-1)^n$ does not exist.

Cheers!

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