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

Given the sequences (u_n )_(n∈N) and (v_n )_(n∈N) defined by u_n =Σ_(k=0) ^n (1/(k!)) and v_n =u_n +(1/(n(n!))) a\ Show that (u_n )_n is of Cauchy. Deduce that (u_n )_n converges. b\ Show that (u_n )_n and (v_n )_n are adjacent c\ Show that their common limit is not a rational number.

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

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

for ∀k≥2 k!≥k(k−1) U_(n+m) −U_n =Σ_(k=1) ^m U_(k+n) ≤Σ_(k=1) ^m (1/((k+n)(k+n−1)))=Σ_(k=1) ^m ((1/(n+k−1))−(1/(n+k))) =(1/m)−(1/(n+m))≤(1/m)....1 U_n cauchy ⇔∀ε>0 ∃N ,∀n,m ≥N ∣U_n −U_m ∣<ε we choose N=E((1/ε))+1⇒U_n −U_m ≤(1/m)≤(1/N)<ε ....by1 U_n is cauchy since is reel sequence ⇒U_n converge b) U_(n+1) −U_n =(1/((n+1)!))>0 U_n is ibcreasing V_(n+1) −V_n =(1/((n+1)!))+(1/((n+1)(n+1)!))−(1/(n.n!)) ((n+2−(n+1)(n+1))/((n+1)(n+1)!))=((−n^2 −n+1)/((n+1)(n+1)!))<0,∀n≥1 V_n decreasing U_n −V_n =(1/(nn!))→0 U_n ,Vn are adjacente if x=(a/b)= lim U_n ∈IQ Σ_(n=0) ^(+∞) (1/(n!))=(a/b)=((a(b−1)!)/(b!))=Σ_(k≥0) (1/(k!)) ⇔a(b−1)!=b!.(Σ_(k=0) ^b (1/(k!)))+(1/(b+1))+Σ_(k≥2) ((b!)/((b+k)!)) k≥2⇒((b!)/((b+k)!))≤(1/((b+k)(b+k−1))) ⇒Σ_(k≥2) ((b!)/((b+k)!))≤Σ_(k≥2) ((1/((b+k)(b+k−1))))=(1/(b+1)) ⇒∣a(b−1)!−b!.Σ_(k=0) ^b (1/(k!))∣≤(2/(b+1))<1 if b≥2....E but a(b−1)! ∈N and b!.Σ_(k=0) ^b (1/(k!))∈N ⇒∣a(b−1)!−b!.Σ_(k=0) ^b (1/(k!))∣∈N⇒Eabsurd ⇒x∉Q

$${for}\:\:\forall{k}\geqslant\mathrm{2}\:\:\:{k}!\geqslant{k}\left({k}−\mathrm{1}\right) \\ $$$${U}_{{n}+{m}} −{U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{U}_{{k}+{n}} \leqslant\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\mathrm{1}}{\left({k}+{n}\right)\left({k}+{n}−\mathrm{1}\right)}=\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\left(\frac{\mathrm{1}}{{n}+{k}−\mathrm{1}}−\frac{\mathrm{1}}{{n}+{k}}\right) \\ $$$$=\frac{\mathrm{1}}{{m}}−\frac{\mathrm{1}}{{n}+{m}}\leqslant\frac{\mathrm{1}}{{m}}....\mathrm{1} \\ $$$${U}_{{n}} \:{cauchy}\:\Leftrightarrow\forall\epsilon>\mathrm{0}\:\exists{N}\:,\forall{n},{m}\:\geqslant{N}\:\mid{U}_{{n}} −{U}_{{m}} \mid<\epsilon \\ $$$${we}\:{choose}\:{N}={E}\left(\frac{\mathrm{1}}{\epsilon}\right)+\mathrm{1}\Rightarrow{U}_{{n}} −{U}_{{m}} \leqslant\frac{\mathrm{1}}{{m}}\leqslant\frac{\mathrm{1}}{{N}}<\epsilon\:\:....{by}\mathrm{1} \\ $$$${U}_{{n}} {is}\:{cauchy}\:{since}\:{is}\:{reel}\:{sequence}\:\Rightarrow{U}_{{n}} \:\:{converge} \\ $$$$\left.{b}\right)\:{U}_{{n}+\mathrm{1}} −{U}_{{n}} =\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}>\mathrm{0}\:\:{U}_{{n}} \:\:{is}\:{ibcreasing} \\ $$$${V}_{{n}+\mathrm{1}} −{V}_{{n}} =\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{{n}.{n}!} \\ $$$$\frac{{n}+\mathrm{2}−\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)!}=\frac{−{n}^{\mathrm{2}} −{n}+\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)!}<\mathrm{0},\forall{n}\geqslant\mathrm{1}\:{V}_{{n}} \:{decreasing} \\ $$$${U}_{{n}} −{V}_{{n}} =\frac{\mathrm{1}}{{nn}!}\rightarrow\mathrm{0}\:\:{U}_{{n}} ,{Vn}\:{are}\:{adjacente} \\ $$$${if}\:{x}=\frac{{a}}{{b}}=\:{lim}\:{U}_{{n}} \in\mathbb{IQ}\:\: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}!}=\frac{{a}}{{b}}=\frac{{a}\left({b}−\mathrm{1}\right)!}{{b}!}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!} \\ $$$$\Leftrightarrow{a}\left({b}−\mathrm{1}\right)!={b}!.\left(\underset{{k}=\mathrm{0}} {\overset{{b}} {\sum}}\frac{\mathrm{1}}{{k}!}\right)+\frac{\mathrm{1}}{{b}+\mathrm{1}}+\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{b}!}{\left({b}+{k}\right)!} \\ $$$${k}\geqslant\mathrm{2}\Rightarrow\frac{{b}!}{\left({b}+{k}\right)!}\leqslant\frac{\mathrm{1}}{\left({b}+{k}\right)\left({b}+{k}−\mathrm{1}\right)} \\ $$$$\Rightarrow\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{b}!}{\left({b}+{k}\right)!}\leqslant\underset{{k}\geqslant\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{\left({b}+{k}\right)\left({b}+{k}−\mathrm{1}\right)}\right)=\frac{\mathrm{1}}{{b}+\mathrm{1}} \\ $$$$\Rightarrow\mid{a}\left({b}−\mathrm{1}\right)!−{b}!.\underset{{k}=\mathrm{0}} {\overset{{b}} {\sum}}\frac{\mathrm{1}}{{k}!}\mid\leqslant\frac{\mathrm{2}}{{b}+\mathrm{1}}<\mathrm{1}\:\:{if}\:{b}\geqslant\mathrm{2}....{E} \\ $$$${but}\:{a}\left({b}−\mathrm{1}\right)!\:\in\mathbb{N}\:{and}\:{b}!.\underset{{k}=\mathrm{0}} {\overset{{b}} {\sum}}\frac{\mathrm{1}}{{k}!}\in\mathbb{N} \\ $$$$\Rightarrow\mid{a}\left({b}−\mathrm{1}\right)!−{b}!.\underset{{k}=\mathrm{0}} {\overset{{b}} {\sum}}\frac{\mathrm{1}}{{k}!}\mid\in\mathbb{N}\Rightarrow{Eabsurd} \\ $$$$\Rightarrow{x}\notin\mathbb{Q} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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

Thanks for your method,

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

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