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\( x_1=2\ \ \wedge\ \ {x_{n+1}=\sqrt {x_n+\frac 1n}\ ,\ (\forall )\ n\in\mathbb{N}^* \) \( \Longrightarrow \) \( \lim_{n\to\infty}\ x_n= 1 \ \ \wedge\ \ \lim_{n\to\infty}\ x_n^n=e\ . \)

Dem. Se arata prin calcule "istovitoare" ca \( \frac 54\ <\ x_4=\sqrt {\sqrt {\sqrt 3+\frac 12}+\frac 13}\ <\ \frac 32 \)

si pentru orice \( n\ge 4 \) avem \( 1+\frac 1n<x_n<1+\frac {1}{n-2}\ \Longrightarrow\ 1+\frac {1}{n+1}<x_{n+1}<1+\frac {1}{n-1} \) deoarece

\( \left\{\begin{array}{ccccc}
\underline{\overline{\left\|\ 1+\frac 1n\ <\ x_n\ \right\|}} & \Longrightarrow & 1+\frac {1}{n+1}\stackrel {(1)}{\ <\ }\sqrt {1+\frac 1n+\frac 1n}\ <\ \sqrt {x_n+\frac 1n}=x_{n+1} & \Longrightarrow & \underline{\overline{\left\|\ 1+\frac {1}{n+1}\ <\ x_{n+1}\ \right\|}}\\\\\\\\
\underline{\overline{\left\|\ x_n\ <\ 1+\frac {1}{n-2}\ \right\|}} & \Longrightarrow & x_{n+1}=\sqrt {x_n+\frac 1n}\ <\ \sqrt {1+\frac {1}{n-2}+\frac 1n}\stackrel{(2)}{\ <\ }1+\frac {1}{n-1} & \Longrightarrow & \underline{\overline{\left\|\ x_{n+1}\ <\ 1+\frac {1}{n-1}\ \right\|}}\end{array}\ \right\|\ . \)

Se arata usor ca inegalitatea \( (1)\ \Longleftrightarrow\ 0\ <\ 1 \) si inegalitatea \( (2)\ \Longleftrightarrow\ n(n-4)+2\ >\ 0\ . \)

Conform principiului inductiei rezulta \( \overline{\underline{\left\|\ 1+\frac 1n\ <\ x_n\ <\ 1+\frac {1}{n-2}\ ,\ (\forall )\ n\ \ge\ 4\ \right\|}}\ . \) In concluzie, \( x_n\rightarrow 1\ . \)

Din lantul \( 1\ <\ n\left(x_n-1\right)\ <\ \frac {n}{n-2}\ ,\ (\forall )\ n\ \ge\ 4 \) se obtine \( \lim_{n\to\infty}\ x_n^n\stackrel{\left(1^{\infty}\right)}{\ \ \ =\ \ \ }\lim_{n\to\infty}\ e^{n\left(x_n-1\right)}=e\ . \)

Solutia mea din concurs pentru partea mai grea, adica \( \lim_{n \to \infty} x_n =1 \).

Se demonstreaza usor prin inductie ca:

\( x_n \geq 1 \), pentru orice \( n \) natural.

Acum, din inegalitatea mediilor, obtinem:

\( x_{n+1}= \sqrt{x_n + \frac{1}{n}} \leq \frac{x_n+ \frac{1}{n}}{2}+ \frac{1}{2} \)

Repetand acest rationament, gasim:

\( x_{n+1} \leq \frac{x_1}{2^n} + \sum_{k=1}^n {\frac{1}{2^k \cdot (n+1-k)}} + \sum_{k=1}^n \frac{1}{2^k} \).

Adica:

\( 1 \leq \lim_{n \to \infty} x_{n+1} \leq \lim_{n \to \infty} \left(\frac{x_1}{2^n}+ \frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}{2}}-1 +\sum_{k=1}^n {\frac{1}{2^k \cdot (n+1-k)}} \right) \). (1)

Voi demonstra acum ca :
\( \lim_{n \to \infty} a_n = \lim_{n\to \infty} \sum_{k=1}^n {\frac{1}{2^k \cdot (n+1-k)}}=0. \)

Dar din inegalitatea lui Chebysev pentru sirurile invers ordonate, obtinem:

\( \sum_{k=1}^n {\frac{1}{2^k \cdot (n+1-k)}} \leq \frac{H_n}{n} \left( 1- \frac{1}{2^n}\right) \to 0 \) si concluzia este evidenta.

Acum aplicand clestele in relatia (1), obtinem ca \( \lim_{n \to \infty}{x_n}=1 \).

Obs: Am notat cu \( H_n= \sum_{k=1}^n{1+ \frac{1}{2}+ \dots + \frac{1}{n}} \).

Solutia mea din concurs pentru \( \lim_{n\to\infty}x_n=1 \).

Se demonstreaza usor prin inductie ca \( x_n \geq 1 \).

Acum aratam prin inductie ca \( x_n\leq 1+ \frac{2}{n} \)
P(1): \( x_1 \leq 3 \) adevarata

Presupunem P(k) adevarata si dem. \( P(k)\to P(k+1) \).
Asadar \( x_{k+1}=\sqrt{x_k+\frac{1}{k}}\leq\sqrt{1+\frac{3}{k}} \).
Prin calcul iese imediat ca \( \sqrt{1+\frac{3}{k}}\leq 1+\frac{2}{k+1} \),
deci \( x_{k+1}\leq 1+\frac{2}{k+1} \).

\( P(k) \) implica \( P(k+1) \) si P(1) adevarata, deci P(n) adevarata \( \forall n\in \mathbb{N} \).

Deci \( 1\leq x_n \leq 1+\frac{2}{n} \).

Din Teorema Clestelui ajungem la concluzia cautata.

Deoarece Laurian Filip si Turcas au prezentat doua solutii pentru \( x_n\rightarrow 1 \) fara a cerceta monotonia sirului \( (x_n) \) ,
voi oferi si o solutie bazata pe acest fapt. Se arata usor inductiv ca pentru orice \( n\in\mathbb{N}^* \) avem \( x_n>1 \).
Deoarece \( 2=x_1>x_2=\sqrt 3 \) si \( x_n>x_{n+1} \) \( \Longrightarrow \) \( x_n+\frac 1n>x_{n+1}+\frac {1}{n+1} \) \( \Longrightarrow \) \( x^2_{n+1}>x^2_{n+2} \) \( \Longrightarrow \)
\( x_{n+1}>x_{n+2} \), ceea ce inseamna ca sirul \( (x_n) \) este strict descrescator. Deoarece este marginit la stanga de \( 1 \) ,
rezulta ca este convergent, fie \( x_n\rightarrow l\ge 1 \). Pentru \( n\rightarrow\infty \) in relatia de recurenta se obtine \( l^2=l\ge 1 \), adica \( l=1 \).