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Let us define a series $\{a(n)\}$ such that $a(n)= a(n-1) + \frac{1}{a(n-1)}$ and $a(1)=1$

then prove that $a(75)$ belongs to the interval $(12,15)$

note here i have used $X(n)$ where n is in sub script denoting the series count.

Im unable to think.. what i have done is i have added all the terms as $\{ a(n)-a(1)\}$ which gave me: $a(n)-a(1)=\frac{1}{a(1)} +\frac{1}{a(2)}+\frac{1}{a(3)}......+\frac{1}{a(n-1)}$

i tried am-gm inequality but it is of no help.

Please Help .

User
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Abhash Jha
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1 Answers1

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Notice first that

$$ a_n^2 = a_{n-1}^2 + 2 + \frac{1}{a_{n-1}^2} > a_{n-1}^2 + 2. \tag{*}$$

Recursively applying this inequality and using the fact that $a_2 = 2$,

$$ \forall n \geq 2, \qquad a_n^2 \geq 2n \tag{1} $$

Now plugging this inequality back to $\text{(*)}$, for $n \geq 3$ we have

$$ a_n^2 \leq a_{n-1}^2 + 2 + \frac{1}{2(n-1)} $$

with strict inequality when $n \geq 3$. Recursively applying this inequality yields

$$ a_n^2 \leq a_2^2 + \sum_{k=2}^{n-1} \left(2 + \frac{1}{2k} \right) = 2n + \sum_{k=2}^{n-1} \frac{1}{2k} \tag{2} $$

When $n = 75$, we obtain a lower bound from $\text{(1)}$:

$$ 12^2 < 150 \leq a_{75}^2 $$

and an upper bound from $\text{(2)}$:

$$ a_{75}^2 \leq 150 + \sum_{k=2}^{74} \frac{1}{2k} < 150 + \frac{\log 74}{2} < 150 + \frac{\log_2 128}{2} < 169 = 13^2. $$

This proves that $ 12 < a_{75} < 13 $ and hence the claim follows.

Sangchul Lee
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    It is enough for the upper bound to use $a_n^2=a_{n-1}^2+2+1/a_{n-1}^2<a_{n-1}^2+3$ so $a_n^2<3n$. – guest Apr 14 '17 at 06:14
  • @guest, You are right. Indeed, the interval is cooked up so that even such a crude bound works to yield a proof. – Sangchul Lee Apr 14 '17 at 06:20