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Question Number 148635 by learner001 last updated on 29/Jul/21
show that {a_n }:=(1/(1!))−(1/(2!))+(1/(3!))−(1/(4!))...+(((−1)^(n+1) )/(n!)) is a cauchy sequence. my attempt: let ε>0 we have ∣a_m −a_n ∣=∣(((−1)^(n+2) )/((n+1)!))+(((−1)^(n+3) )/((n+2)!))+...+(((−1)^(m+1) )/(m!))∣ ≤∣(((−1)^(n+2) )/((n+1)!))∣+∣(((−1)^(n+3) )/((n+2)!))∣+...+∣(((−1)^(m+1) )/(m!))∣ =(1/((n+1)!))+(1/((n+2)!))+...+(1/(m!))<(1/(n!))≤(1/n)<ε. (1/n)→0 as n→∞, so no matter any ε>0 (1/n)<ε eventually. as long as n^∗ >(1/ε) ∣a_m −a_n ∣<ε ∀n≥n^∗ .
$$\mathrm{show}\:\mathrm{that}\:\left\{\mathrm{a}_{\mathrm{n}} \right\}:=\frac{\mathrm{1}}{\mathrm{1}!}−\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}…+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}!}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cauchy} \\ $$$$\mathrm{sequence}. \\ $$$$\mathrm{my}\:\mathrm{attempt}: \\ $$$$\mathrm{let}\:\epsilon>\mathrm{0}\:\mathrm{we}\:\mathrm{have}\:\mid\mathrm{a}_{\mathrm{m}} −\mathrm{a}_{\mathrm{n}} \mid=\mid\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{2}} }{\left(\mathrm{n}+\mathrm{1}\right)!}+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{3}} }{\left(\mathrm{n}+\mathrm{2}\right)!}+…+\frac{\left(−\mathrm{1}\right)^{\mathrm{m}+\mathrm{1}} }{\mathrm{m}!}\mid \\ $$$$\leqslant\mid\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{2}} }{\left(\mathrm{n}+\mathrm{1}\right)!}\mid+\mid\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{3}} }{\left(\mathrm{n}+\mathrm{2}\right)!}\mid+…+\mid\frac{\left(−\mathrm{1}\right)^{\mathrm{m}+\mathrm{1}} }{\mathrm{m}!}\mid \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}+\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)!}+…+\frac{\mathrm{1}}{\mathrm{m}!}<\frac{\mathrm{1}}{\mathrm{n}!}\leqslant\frac{\mathrm{1}}{\mathrm{n}}<\epsilon. \\ $$$$\frac{\mathrm{1}}{\mathrm{n}}\rightarrow\mathrm{0}\:\mathrm{as}\:\mathrm{n}\rightarrow\infty,\:\mathrm{so}\:\mathrm{no}\:\mathrm{matter}\:\mathrm{any}\:\epsilon>\mathrm{0}\:\frac{\mathrm{1}}{\mathrm{n}}<\epsilon\:\mathrm{eventually}. \\ $$$$\mathrm{as}\:\mathrm{long}\:\mathrm{as}\:\mathrm{n}^{\ast} >\frac{\mathrm{1}}{\epsilon}\:\mid\mathrm{a}_{\mathrm{m}} −\mathrm{a}_{\mathrm{n}} \mid<\epsilon\:\forall\mathrm{n}\geqslant\mathrm{n}^{\ast} .\: \\ $$

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