From b685aad981519e40423b7eb8c693fd4b845aa6ca Mon Sep 17 00:00:00 2001
From: Fabian Gabel <fabian.gabel@tuhh.de>
Date: Mon, 13 Jun 2022 16:36:35 +0200
Subject: [PATCH] fix reference
---
nodes/114/114.tex | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/nodes/114/114.tex b/nodes/114/114.tex
index f351ee98..3603dd93 100644
--- a/nodes/114/114.tex
+++ b/nodes/114/114.tex
@@ -46,7 +46,7 @@ Using this inequality, we obtain for $n\geq1$
\[a_{n+1}=\frac{a_n+\frac{2}{a_n}}{2}\geq \sqrt{a_n\cdot\frac2{a_n}}=\sqrt{2}.\]
Thus, $(a_n)$ is bounded from below. For showing monotonicity, we consider
\[a_{n+1}-a_n=\frac{a_n+\frac2{a_n}}{2}-a_n=\frac{1}{2a_n}(2-a_n^2).\]
-In particular, if $n\geq2$, we have that $a_n>0$ and $2-a_n^2\leq0$. Thus, $a_{n+1}-a_n\leq0$ for $n\geq2$. An~application of Theorem~\ref{thm:bndmonseq} (resp.\ the slight generalisation in Remark %\ref{rem:monseqgen}
+In particular, if $n\geq2$, we have that $a_n>0$ and $2-a_n^2\leq0$. Thus, $a_{n+1}-a_n\leq0$ for $n\geq2$. An~application of Theorem~\ref{thm:monbndseq} (resp.\ the slight generalisation in Remark %\ref{rem:monseqgen}
from above)
now leads to the existence of some $a\in\mathbb{R}$ with $a=\lim_{n \to\infty}a_n$.\\[2ex]
To compute the limit, we make use of the relation $\lim_{n \to\infty}a_n=\lim_{n\to\infty}a_{n+1}$ (follows directly from the Definition of limits) and the formulae for limits. This yields
--
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