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For a given sequence $(a_n)$, where $a_1>0$ and for each natural number $n\ge 1$ :

$$a_{n+1}= \dfrac{a_n^2 + 1}{a_n}$$

Prove that a sequence $(a_n)$ diverges.

Proof: (objectionable evidence)

Lower barrier 0, prove. $a_n>0$ $\forall n\in N$. $N$ is natural number and $N>1$.

$n=1$

$a_{1+1}=\frac{a_1\cdot a_1+1}{a_1}=a_1+\frac{1}{a_1} >0$

$a_n>0$ $\Rightarrow a_{n+1}>0$ :

$a_{n+1}=a_n+\frac{1}{a_n}>0$, because $a_n > 0$

Monotonics:

$a_n\geq a_{n+1}$

$a_n-a_{n+1}=a_n-a_n-\frac{1}{a_n} < 0$, because $\frac{1}{a_n}>0$

Sequence is growing.

lim $a_{n+1} = A$

lim $a_{n+1} =$ lim $(a_n+\frac{1}{a_n}) =$ lim $a_n +$ lim $\frac{1}{a_n} = A + \frac{1}{A} = \frac{A\cdot A+1}{A}$ ....

Is this proof correct? Where are mistakes? The correct end of proof?

DeepSea
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Anette Uiga
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1 Answers1

3

$a_{n+1} = a_n + \dfrac{1}{a_n}\implies a_{n+1}^2 = a_n^2 + 2+\dfrac{1}{a_n^2}\implies a_{n+1}^2 - a_n^2 > 2, \forall n \ge 1 \implies a_n^2 = (a_n^2- a_{n-1}^2)+(a_{n-1}^2-a_{n-2}^2)+(a_{n-2}^2-a_{n-3}^2)+\cdots+(a_2^2-a_1^2)+a_1^2> 2(n-1)+a_1^2 > 2(n-1)\implies a_n > \sqrt{2(n-1)} \implies a_n \to \infty$.

DeepSea
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