Question Number 1608 by 112358 last updated on 26/Aug/15
$${Find}\:{the}\:{limit}\:{of}\:{this}\:{sequence}. \\ $$$$\sqrt{\mathrm{2}},\sqrt{\mathrm{2}\sqrt{\mathrm{2}}},\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}},\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}}},... \\ $$$${Show}\:{that}\:{the}\:{sum}\:{of}\:{the} \\ $$$${terms}\:{of}\:{this}\:{infinite}\:{sequence} \\ $$$${does}\:{not}\:{converge}. \\ $$
Commented by Rasheed Ahmad last updated on 26/Aug/15
$${Let}\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}...}}}}={x} \\ $$$${Squaring}\:{both}\:{sides}, \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}...}}}={x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}.{x}={x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{2} \\ $$$${The}\:{limit}\:{of}\:{the}\:{sequence}=\mathrm{2} \\ $$$$ \\ $$
Commented by 112358 last updated on 26/Aug/15
$${Indeed}\:{it}\:{is}!\:{L}=\mathrm{2}\:. \\ $$
Answered by 123456 last updated on 26/Aug/15
$${a}_{{n}} =\sqrt{\mathrm{2}{a}_{{n}−\mathrm{1}} } \\ $$$${a}_{\mathrm{0}} =\sqrt{\mathrm{2}} \\ $$$${a}_{{n}} >\mathrm{0}\Rightarrow\mathrm{2}{a}_{{n}} >\mathrm{0}\Rightarrow\sqrt{\mathrm{2}{a}_{{n}} }>\mathrm{0}\Rightarrow{a}_{{n}+\mathrm{1}} >\mathrm{0} \\ $$$$\mathrm{0}<{a}_{{n}} <\mathrm{2}\Rightarrow\mathrm{0}<\mathrm{2}{a}_{{n}} <\mathrm{4}\Rightarrow\mathrm{0}<\sqrt{\mathrm{2}{a}_{{n}} }<\mathrm{2}\Rightarrow\mathrm{0}<{a}_{{n}+\mathrm{1}} <\mathrm{2} \\ $$$$\mathrm{0}<{a}_{{n}} <{a}_{{n}+\mathrm{1}} <\mathrm{2}\Rightarrow\mathrm{0}<\mathrm{2}{a}_{{n}} <\mathrm{2}{a}_{{n}+\mathrm{1}} <\mathrm{4}\Rightarrow\mathrm{0}<\sqrt{\mathrm{2}{a}_{{n}} }<\sqrt{\mathrm{2}{a}_{{n}+\mathrm{1}} }<\mathrm{2}\Rightarrow\mathrm{0}<{a}_{{n}+\mathrm{1}} <{a}_{{n}+\mathrm{2}} <\mathrm{2} \\ $$$$\mathrm{0}<{a}_{{n}+\mathrm{1}} <{a}_{{n}} <\mathrm{2}\Rightarrow\mathrm{0}<\mathrm{2}{a}_{{n}+\mathrm{1}} <\mathrm{2}{a}_{{n}} <\mathrm{4}\Rightarrow\mathrm{0}<\sqrt{\mathrm{2}{a}_{{n}+\mathrm{1}} }<\sqrt{\mathrm{2}{a}_{{n}} }<\mathrm{2}\Rightarrow\mathrm{0}<{a}_{{n}+\mathrm{2}} <{a}_{{n}+\mathrm{1}} <\mathrm{2} \\ $$$${a}_{\mathrm{0}} =\sqrt{\mathrm{2}},{a}_{\mathrm{1}} =\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}\Rightarrow{a}_{\mathrm{1}} >{a}_{\mathrm{0}} ,\:\mathrm{the}\:{a}\:\mathrm{is}\:\mathrm{crescent} \\ $$$$\mathrm{since}\:{a}\:\mathrm{is}\:\mathrm{bounced}\:\mathrm{and}\:\mathrm{crescent},\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{a}_{{n}} \:\mathrm{exists} \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{{n}} ={a} \\ $$$${a}=\sqrt{\mathrm{2}{a}}\Rightarrow{a}^{\mathrm{2}} =\mathrm{2}{a}\Rightarrow{a}\left({a}−\mathrm{2}\right)=\mathrm{0}\Rightarrow{a}=\mathrm{0}\vee{a}=\mathrm{2} \\ $$$$\mathrm{since}\:\mathrm{it}\:\mathrm{crescent}\:\mathrm{them}\:{a}=\mathrm{2},\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{a}_{{n}} ={a}=\mathrm{2} \\ $$$$\mathrm{S}=\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}{a}_{{n}} \\ $$$$\mathrm{since}\:\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\:{a}_{{n}} \neq\mathrm{0},\:\mathrm{then}\:\mathrm{the}\:\mathrm{serie}\:\mathrm{diverge} \\ $$
Commented by Rasheed Ahmad last updated on 27/Aug/15
$${What}\:{is}\:{meant}\:{by}\:'{crescent}'? \\ $$$${I}\:{am}\:{asking}\:{for}\:{the}\:{sake}\:{of} \\ $$$${knowledge}. \\ $$
Commented by 123456 last updated on 27/Aug/15
$$\mathrm{growing}\:\mathrm{or}\:\mathrm{something}\:\mathrm{like}\:\mathrm{this}\:\left(\mathrm{sorry}\:\mathrm{about}\:\mathrm{my}\:\mathrm{bad}\:\mathrm{english}\right) \\ $$$$\mathrm{as}\:\mathrm{shown} \\ $$$$\mathrm{0}<{a}_{{n}} <{a}_{{n}+\mathrm{1}} <\mathrm{2}\Rightarrow\mathrm{0}<{a}_{{n}+\mathrm{1}} <{a}_{{n}+\mathrm{2}} <\mathrm{2} \\ $$$$\mathrm{since}\:\mathrm{0}<{a}_{\mathrm{0}} <{a}_{\mathrm{1}} <\mathrm{2},\:\mathrm{them}\:\left(\mathrm{0}<\mathrm{2}<\mathrm{2}\sqrt{\mathrm{2}}<\mathrm{4}\Rightarrow\mathrm{0}<\sqrt{\mathrm{2}}<\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}<\mathrm{2}\right) \\ $$$$\mathrm{0}<{a}_{\mathrm{0}} <{a}_{\mathrm{1}} <{a}_{\mathrm{2}} <{a}_{\mathrm{3}} <\centerdot\centerdot\centerdot<\mathrm{2} \\ $$
Commented by 112358 last updated on 27/Aug/15
![He/she′s saying that the terms of the sequence are perpetually increasing to a limiting value of 2. Hence, the sequence is monotonic and upper bounded. i.e a_n ≤M for all n and a_(n+1) ≥a_n for all n. Generally, all sequences which are monotonic (only increasing or decreasing but not both) and bounded (upper or lower[a_n >M for any finite n]) are convergent. This is called the monotonic sequence theorem.](Q1617.png)
$${He}/{she}'{s}\:{saying}\:{that}\:{the}\:{terms}\: \\ $$$${of}\:{the}\:{sequence}\:{are}\:{perpetually} \\ $$$${increasing}\:{to}\:{a}\:{limiting}\:{value}\:{of}\: \\ $$$$\mathrm{2}.\:{Hence},\:{the}\:{sequence}\:{is} \\ $$$${monotonic}\:{and}\:{upper}\:{bounded}. \\ $$$${i}.{e}\:\:{a}_{{n}} \leqslant{M}\:{for}\:{all}\:\:{n}\:\:{and} \\ $$$${a}_{{n}+\mathrm{1}} \geqslant{a}_{{n}} \:{for}\:{all}\:\:{n}. \\ $$$${Generally},\:{all}\:{sequences}\:{which} \\ $$$${are}\:{monotonic}\:\left({only}\:{increasing}\right. \\ $$$$\left.{or}\:{decreasing}\:{but}\:{not}\:{both}\right)\:{and}\:{bounded} \\ $$$$\left({upper}\:{or}\:{lower}\left[{a}_{{n}} >{M}\:\:{for}\:{any}\right.\right. \\ $$$$\left.{f}\left.{inite}\:{n}\right]\right)\:{are}\:{convergent}.\:{This} \\ $$$${is}\:{called}\:{the}\:{monotonic}\: \\ $$$${sequence}\:{theorem}. \\ $$
Commented by Rasheed Ahmad last updated on 27/Aug/15
$$\boldsymbol{\mathrm{Thanks}}\:\mathrm{to}\:\mathrm{both}\:\mathrm{of}\:\mathrm{you}! \\ $$