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Question Number 97781 by Ar Brandon last updated on 09/Jun/20

$$\mathrm{Given}\mathrm{the}\mathrm{sequences}{\left({\mathrm{u}}_{\mathrm{n}}\right)}_{\mathrm{n}\in \mathbb{N}}\mathrm{and}{\left({\mathrm{v}}_{\mathrm{n}}\right)}_{\mathrm{n}\in \mathbb{N}}\mathrm{defined}$$$$\mathrm{by}{\mathrm{u}}_{\mathrm{n}}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{k}!}\mathrm{and}{\mathrm{v}}_{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}+\frac{1}{\mathrm{n}(\mathrm{n}!)}$$$$\mathrm{a}\mathrm{\setminus }\mathrm{Show}\mathrm{that}{\left({\mathrm{u}}_{\mathrm{n}}\right)}_{\mathrm{n}}\mathrm{is}\mathrm{of}\mathrm{Cauchy}.\mathcal{D}\mathrm{educe}\mathrm{that}$$$${\left({\mathrm{u}}_{\mathrm{n}}\right)}_{\mathrm{n}}\mathrm{converges}.$$$$\mathrm{b}\mathrm{\setminus }\mathrm{Show}\mathrm{that}{\left({\mathrm{u}}_{\mathrm{n}}\right)}_{\mathrm{n}}\mathrm{and}{\left({\mathrm{v}}_{\mathrm{n}}\right)}_{\mathrm{n}}\mathrm{are}\mathrm{adjacent}$$$$\mathrm{c}\mathrm{\setminus }\mathrm{Show}\mathrm{that}\mathrm{their}\mathrm{common}\mathrm{limit}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{rational}$$$$\mathrm{number}.$$

Answered by maths mind last updated on 10/Jun/20

$$for\mathrm{\forall }k⩾2k!⩾k(k-1)$$$$U_{n+m}-U_n=\underset{k=1}{\sum ^m}{U}_{k+n}⩽\underset{k=1}{\sum ^m}\frac{1}{(k+n)(k+n-1)}=\underset{k=1}{\sum ^m}(\frac{1}{n+k-1}-\frac{1}{n+k})$$$$=\frac{1}{m}-\frac{1}{n+m}⩽\frac{1}{m}....1$$$$U_{n}cauchy\iff \mathrm{\forall }ϵ>0\mathrm{\exists }N,\mathrm{\forall }n,m⩾N\mid U_n-U_m\mid <ϵ$$$$wechooseN=E\left(\frac{1}{ϵ}\right)+1\Rightarrow U_n-U_m⩽\frac{1}{m}⩽\frac{1}{N}<ϵ....by1$$$$U_{n}iscauchysinceisreelsequence\Rightarrow U_{n}converge$$$$b)U_{n+1}-U_n=\frac{1}{(n+1)!}>0U_{n}isibcreasing$$$$V_{n+1}-V_n=\frac{1}{(n+1)!}+\frac{1}{(n+1)(n+1)!}-\frac{1}{n.n!}$$$$\frac{n+2-(n+1)(n+1)}{(n+1)(n+1)!}=\frac{-n^2-n+1}{(n+1)(n+1)!}<0,\mathrm{\forall }n⩾1V_{n}decreasing$$$$U_n-V_n=\frac{1}{nn!}\to 0U_n,Vnareadjacente$$$$ifx=\frac{a}{b}=lim{U}_n\in \mathbb{IQ}$$$$\underset{n=0}{\sum ^{+\mathrm{\infty }}}\frac{1}{n!}=\frac{a}{b}=\frac{a(b-1)!}{b!}=\sum _{k⩾0}\frac{1}{k!}$$$$\iff a(b-1)!=b!.\left(\underset{k=0}{\sum ^b}\frac{1}{k!}\right)+\frac{1}{b+1}+\sum _{k⩾2}\frac{b!}{(b+k)!}$$$$k⩾2\Rightarrow \frac{b!}{(b+k)!}⩽\frac{1}{(b+k)(b+k-1)}$$$$\Rightarrow \sum _{k⩾2}\frac{b!}{(b+k)!}⩽\sum _{k⩾2}\left(\frac{1}{(b+k)(b+k-1)}\right)=\frac{1}{b+1}$$$$\Rightarrow \mid a(b-1)!-b!.\underset{k=0}{\sum ^b}\frac{1}{k!}\mid ⩽\frac{2}{b+1}<1ifb⩾2....E$$$$buta(b-1)!\in \mathbb{N}andb!.\underset{k=0}{\sum ^b}\frac{1}{k!}\in \mathbb{N}$$$$\Rightarrow \mid a(b-1)!-b!.\underset{k=0}{\sum ^b}\frac{1}{k!}\mid \in \mathbb{N}\Rightarrow Eabsurd$$$$\Rightarrow x\notin \mathbb{Q}$$$$$$$$$$$$$$

Commented by Ar Brandon last updated on 10/Jun/20

$$\mathcal{T}\mathrm{hanks}\mathrm{for}\mathrm{your}\mathrm{method},$$

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