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We just had our first calculus lecture, and I'm kinda stuck at this proof right now:

Prove that $\lim_{n\to\infty}\{(-1)^n\}$ diverges.

Given: $\lim_{n\to\infty}(-1)^n = a$. Thus for $\varepsilon_0 = 1 $ there should be $n_0 \in\mathbb N$ so that $\forall n > n_0; |(-1)^n-a| < 1 $ But when $n > n_0$, then $|(-1)^n-(-1)^{n+1}| \le |(-1)^n-a|+|a-(-1)^{n+1}| < 1+1 = 2$ -> A contradiction!

Now I think I lose it at $|(-1)^n-(-1)^{n+1}|$ Why does he substitute 'a' with $(-1)^{n+1}$, or does he anyways?

any help is appreciated !

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    Here is a fun problem. If $a_n$ is a convergent sequence, then the sequence $|a_{n+1}-a_n|$ converges to zero.2012-09-12

4 Answers 4

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This is not a substitution; it's an application of the triangle inequality $|a-b|\le|a-c|+|c-b|$.

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    i think i just got that :D2012-09-12
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You know that there is a theorem stating:

If $\{a_n\}$ converges then every subsequence of it converges.

Therefore, if you can find two subsequences of a sequence $\{a_n\}$ which converge to the different limits, then the sequence $\{a_n\}$ does not converge! Here are two converging subsequences $1, 1, 1, 1, ...$ and $-1, -1, -1, -1, ...$ which converge to $1$ and $-1$ respectively. And, $1\neq -1$.

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    I like this; another "must know" theorem! $\quad\ddot\smile\quad$2013-03-17
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Recall the definition of convergence. For every $\epsilon$ their is an $n_0$ such that FOR ALL $n>n_0$, $|a_n-a|<\epsilon$. Now Pick $\epsilon=1$ (what you called $\epsilon_0$). Then for some $n_0$, $|a_n-a|<1$ (since 1 is our $\epsilon$). Thus $n+1$ is greater than $n$. This is why you instructor may make that substitution. $1=|(-1)^n-(-1)^{n+1}| \leq |(-1)^n-a|+|a-(-1)^{n+1}| < 1+1 = 2$ Note that the first inequality is just the triangle inequality. Now for the second. If $n>n_0$ ($n_0$ being the n-value that makes $\epsilon=1$), then $n+1>n>n_0$. Since both numbers are greater than $n_0$, we get the two inequalities, $|(-1)^n-a|,|a-(-1)^{n+1}|<1 $.

At the risk of being redundant let me again say why your instructor may use $n+1$. It is because if $n_0$ is the n-value that make $\epsilon=1$, and $n>n_0$, then $|a_n-a|<\epsilon = 1$. But as $n+1$ is also greater than $n_0$, then $|a_{n+1}-a|<\epsilon =1 $ , as well.

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    Let us look at the definition of convergence. Defn: For every \epsilon >0 their exists $n_0$ such that for all n>n_0, |a_n-a|<\epsilon. So let us start withe the first piece: It states that For every $\epsilon$ their is an $n_0$ such that STUFF HAPPENS (we will suppress that for now). So what I mean is that if you have some $\epsilon$ (in your case $\epsilon =1$) then their is some $n_0$. So what I meant by that phrase is that $n_0$ will be dependent upon the chosen $\epsilon$.2012-09-12
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By the definition of limit, calling $p_n=(-1)^n$, you know that $p_n$ is as close as you want to its limit, definitely. When $n$ is even, $p_n=1$, and it must be as close as you want to a fixed number. This number must be $1$. By the same token, when $n$ is odd, $p_n=-1$ must be as close as you want to a fixed number, and therefore this number must be $-1$. Since limits are unique, you conclude that no limit can exist for $\{p_n\}_n$.

It is just a reformulation of the "subsequence principle" cited by @BabakSorouh in his answer, but in this particular case it can be shown by hands.