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Prove: If $\lim_{n\to\infty}a_n = L$ and $a_n > a$ for all $n$ then $L \geq a$

Proof: We know from the definition of the limit that $\forall_{\epsilon > 0} \exists_n s.t. \forall_{n>N} |a_n - L| < \epsilon$. Now since $a_n > a$ for all $n$...

I am not really sure where to go from here. Is it the case that all sequences defined by this statement are monotone non-increasing? Then intuitively we could say $a_n = L$ for sufficiently large $n$. Thus, by transitivity $L > a$

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Suppose that $L<a$. We put $\varepsilon=a-L>0$. Since $\displaystyle\lim_{n\rightarrow\infty}a_n=L$, there exists $N_0\in \mathbb{N}$ such that $$ |a_n-L|<\varepsilon \quad\forall n\geq N_0. $$ Then $a_n-L<a-L$ for all $n\geq N_0$, or $a_n<L$ for all $n\geq N_0$. This is contradict to the assumption $a_n>a$ for all $n$. Hence $L\geq a$. In the above argument, we have seen that we only need $a_n>L$ for sufficiently large $n$.

Moreover, in the general case we do not have $L>a$. Indeed, we observe that although $\displaystyle a_n=\frac{1}{n}>0$ for all $n$ but $\displaystyle L=\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}\frac{1}{n}=0$.

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  • $\begingroup$ @CodeKingPlusPlus: Please see the solution and extended disscusion on your question. $\endgroup$
    – blindman
    Oct 5 '12 at 4:56
  • $\begingroup$ Where and what is the extended discussion? $\endgroup$ Oct 5 '12 at 14:10
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Hint: To show that $L\geqslant a$, it is enough to show that $a\leqslant L+\varepsilon$, for every $\varepsilon\gt0$. Now, $a_n\to L$ hence, for every $\varepsilon\gt0$, there exists $n_\varepsilon$ such that for every $n\geqslant n_\varepsilon$, $a_n$ is such that...

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  • $\begingroup$ Why is it enough to show that $a \leq L +\epsilon,$ for every $\epsilon > 0$? If it is, that is what I have. $\endgroup$ Sep 29 '12 at 20:05
  • $\begingroup$ Suppose by contradiction that $L<a\leq L+\epsilon \ \forall \epsilon$... $\endgroup$ Sep 29 '12 at 20:30
  • $\begingroup$ @PlusPlus If $a\leqslant L+\varepsilon$ for every $\varepsilon\gt0$, then $a\leqslant\inf\{L+\varepsilon\mid\varepsilon\gt0\}=\ldots$ $\endgroup$
    – Did
    Sep 29 '12 at 20:54
  • $\begingroup$ Does that mean we take $\epsilon$ to be practically zero? Where did you get that implication $a \leq inf\{L+\epsilon|\epsilon >0\}$ $\endgroup$ Oct 1 '12 at 0:03
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    $\begingroup$ For any set $S$, [$a\leqslant s$ for every $s$ in $S$] is logically equivalent to [$a\leqslant\inf S$]. Is this your question? $\endgroup$
    – Did
    Oct 1 '12 at 5:32
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The sequence need not be monotone. But, here's a hint for one approach towards a proof:

Try arguing by contradiction: Assume $L<a$. Now choose an $\epsilon>0$ so that $L+\epsilon<a$. What can you say about the terms $a_n$ of the sequence for large $n$?

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  • $\begingroup$ my intuition tells me that for large values of $n$ that I can say: $L = a$ and then $a + \epsilon$ is not less than $a$ $\endgroup$ Sep 29 '12 at 22:19
  • $\begingroup$ @CodeKingPlusPlus For large $n$, you would know $|a_n-L|<\epsilon$; and this would imply $a_n<L+\epsilon$. But $L+\epsilon<a$, so... $\endgroup$ Sep 29 '12 at 23:20

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