a_(n+1) =a_n ^2 +2a_n −2, a_1 =3. Prove thatΣ (1/(a


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Question Number 203313 by CrispyXYZ last updated on 16/Jan/24

$a_{n + 1} = a_{n}^{2} + 2 a_{n} - 2, a_{1} = 3 .$ $Prove that Σ \frac{1}{a_{n} + 2} ⩽ \frac{3}{10} .$
	Answered by witcher3 last updated on 16/Jan/24
	${ ((a_(n+1) =a_n ^2 +2a_n −2)),((a_1 =3)) :} a_n ≥n+2 by recursion a_1 =3≥1+2 true suppose ∀n≥1 a_n ≥n+2 a_(n+1) =f(a_n );f(x)=x^2 +2x−2 increse in [0,∞[ a_n ≥n+2>0⇒a_(n+1) =f(a_n )≥f(n+2)=(n+2)^2 +2(n+2)−2 =n^2 +6n+6>n+3 ⇒a_(n+1) >n+3 ⇒∀n∈N a_n ≥n+2 a_(n+1) =a_n ^2 +2a_n −2⇒a_(n+1) +2=a_n (a_n +2) ⇒((a_(n+1) +2)/(a_n +2))=a_n ≥n+2 ⇒Π_(k=1) ^n ((a_(k+1) +2)/(a_k +2))≥Π_(k=1) ^n (k+2) ⇒((a_(n+1) +2)/(a_1 +2))≥3.4....(n+2)=(((n+2)!)/2) ⇒a_(n+1) +2≥(5/2)(n+2)! a_n +2≥(5/2)(n+1)!;∀n∈N ⇒(1/(a_n +2))≤(2/5).(1/((n+1)!))⇒Σ_(n≥1) (1/(a_n +2))≤(2/5)Σ_(n≥1) (1/((n+1)!))=(2/5)(Σ_(n≥0) (1/(n!))−2) e=Σ_(n≥0) (1/(n!)) Σ_(n=1) ^∞ (1/(a_n +2))=(2/5)(e−2)<(3/(10)) proof⇔2(e−2)<(3/2)⇒e<2+(3/4)=2.75>e true$
	${\begin{cases} a_{n + 1} = a_{n}^{2} + {2 a}_{n} - 2 \\ a_{1} = 3 \end{cases}$ $a_{n} ⩾ n + 2$ $by recursion a_{1} = 3 ⩾ 1 + 2 true$ $suppose \forall n ⩾ 1 a_{n} ⩾ n + 2$ $a_{n + 1} = f (a_{n}); f (x) = x^{2} + 2 x - 2 increse in [0, \infty [$ $a_{n} ⩾ n + 2 > 0 \Rightarrow a_{n + 1} = f (a_{n}) ⩾ f (n + 2) = {(n + 2)}^{2} + 2 (n + 2) - 2$ $= n^{2} + 6 n + 6 > n + 3$ $\Rightarrow a_{n + 1} > n + 3 \Rightarrow \forall n \in N a_{n} ⩾ n + 2$ $a_{n + 1} = a_{n}^{2} + {2 a}_{n} - 2 \Rightarrow a_{n + 1} + 2 = a_{n} (a_{n} + 2)$ $\Rightarrow \frac{a_{n + 1} + 2}{a_{n} + 2} = a_{n} ⩾ n + 2$ $\Rightarrow \underset{k = 1}{\prod^{n}} \frac{a_{k + 1} + 2}{a_{k} + 2} ⩾ \underset{k = 1}{\prod^{n}} (k + 2)$ $\Rightarrow \frac{a_{n + 1} + 2}{a_{1} + 2} ⩾ 3.4 . . . . (n + 2) = \frac{(n + 2)!}{2}$ $\Rightarrow a_{n + 1} + 2 ⩾ \frac{5}{2} (n + 2)!$ $a_{n} + 2 ⩾ \frac{5}{2} (n + 1)!; \forall n \in N$ $\Rightarrow \frac{1}{a_{n} + 2} ⩽ \frac{2}{5} . \frac{1}{(n + 1)!} \Rightarrow \sum_{n ⩾ 1} \frac{1}{a_{n} + 2} ⩽ \frac{2}{5} \sum_{n ⩾ 1} \frac{1}{(n + 1)!} = \frac{2}{5} (\sum_{n ⩾ 0} \frac{1}{n!} - 2)$ $e = \sum_{n ⩾ 0} \frac{1}{n!}$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{a_{n} + 2} = \frac{2}{5} (e - 2) < \frac{3}{10}$ $proof \Leftrightarrow 2 (e - 2) < \frac{3}{2} \Rightarrow e < 2 + \frac{3}{4} = 2.75 > e true$
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