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}
from above)
now leads to the existence of some $a\in\mathbb{R}$ with $a=\lim_{n o\infty}a_n$.\\[2ex]
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
This relation leads to the equation $2-a^2=0$, i.e., we either have $a=\sqrt{2}$ or $a=-\sqrt{2}$. However, the latter solution cannot be a limit since all sequence elements are positive. Therefore, we have
"label":"Convergence of Bounded MonotonicSequences",
"label":"Bounded Monotonic\nSequences",
"meta":" ",
"content":"If a sequence of real numbers is bounded and monotonic, then it is convergent.",
"notes":"114-snippet.html",
...
...
@@ -1889,6 +1889,31 @@
"target":"104",
"label":"The proof and a lot of applications of the sandwich theorem are only possible because the limit theorems allow for elementary calculations with limits."
},
"101-114":{
"source":"101",
"target":"114",
"label":"If a sequence is bounded and monotonic, then it converges."
},
"102-114":{
"source":"102",
"target":"114",
"label":"Boundedness alone does not give convergence yet, one also needs monotonicity."
},
"106-114":{
"source":"106",
"target":"114",
"label":"The convergence of bounded sequences follows from Dedekind's theorem which guarantees the existence of suprema of bounded sets."
},
"105-114":{
"source":"106",
"target":"114",
"label":"The limit of a monotonic increasing and bounded sequence is precisely the supremum of its function values. An analogous connection holds for the infimum."
},
"013-102":{
"source":"013",
"target":"102",
"label":"The image of a sequence, i.e. the set of all values that the sequence attains, is a bounded set."