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Let $(a_n)$ be a convergent sequence and $M$ a real number such that $a_n ≤ M$ for each $n$. Using the previous question, or otherwise, prove that $\lim_{n\to \infty}a_n≤M$.

I tried the "version" where $a_n > M$ and was able to arrived at a solution but this one seems like a tough nut to crack!

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2 Answers 2

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Let $L$ be the limit of $(a_n)$, and $\varepsilon > 0$. Then, for all $n$ greater than a certain $n_0$, we have: $$\begin{align}-\varepsilon < L-a_n <\varepsilon \\ \Rightarrow L<a_n+\varepsilon\end{align}$$ Since for any $\varepsilon$ there exists an $a_n$ so that the above inequality is satisfied, and each of these $a_n$ is bounded by $M$, we obtain that, for all $\varepsilon > 0$ $$L<M+\varepsilon$$

This means that $L\leq M$ (to prove this, assume $L>M$ and arrive at a contradiction).

Edit: To adress the comment, the first line in the ones you ask about is the definition of $L$ being the limit. Really, the double inequality is just shorthand for the simultaneous inequalities: $$-\varepsilon < L-a_n \\ \text{and} \\L-a_n < \varepsilon$$

Then we just take the second inequality, and add $a_n$ on both sides to get the second line.

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  • $\begingroup$ I do not understand this $$\begin{align}-\varepsilon < L-a_n <\varepsilon \\ \Rightarrow L<a_n+\varepsilon\end{align}$$ $\endgroup$
    – guest
    Mar 28, 2015 at 22:38
  • $\begingroup$ isn't the definition of limit $$a_n-L$$? In your inequality, it goes the other way around which confuses me. $\endgroup$
    – guest
    Mar 28, 2015 at 22:50
  • $\begingroup$ It is easy to prove that L>M: $$L-M>0$$ $$\epsilon >0$$ $$\text{L-M=$\epsilon $}$$ $$L-\epsilon <a_n<L+\epsilon$$ $$M<a_n<2 L-M$$ $$a_n>M$$ $\endgroup$
    – guest
    Mar 28, 2015 at 22:59
  • $\begingroup$ @guest About the order of $a_n - L$, in the definition you'd have $|a_n - L|<\varepsilon$. Since $|a_n-L| = |L-a_n|$, we can also use $|L-a_n| < \varepsilon$. About your second comment, I'm not exactly sure what you mean by "prove L > M", since this is in fact what you want to disprove. Incidentally, you have disproved it by arriving at a contradiction anyways. $\endgroup$
    – GPerez
    Mar 30, 2015 at 18:43
  • $\begingroup$ This was what I was looking for! Cheers $\endgroup$
    – guest
    Mar 31, 2015 at 5:34
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For every $\epsilon>0$ there is $N$ such that $|\lim_na_n-a_n|<\epsilon$.

Then $\lim_na_n-a_n<\epsilon$, for $n>N$.

We deduce that $\lim_na_n<a_n+\epsilon\leq M+\epsilon$.

Since we have obtained that for all $\epsilon>0$, $\lim_na_n<M+\epsilon$, therefore $\lim_na_n\leq M$.

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  • $\begingroup$ Here's my attempt: Let the limit of the sequence $$a_n$$ be a. $$\text{-$\epsilon <|$}a_n\text{-a$|<\epsilon $ $\forall $n$\geq $N}$$ If we let $$\text{$\epsilon $=M-a}$$ $$\text{2a-M$\leq $}a_n\text{$\leq $M}$$ $\endgroup$
    – guest
    Mar 29, 2015 at 0:10

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