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From listing the first few terms, I suspect that the sequence is increasing, so I wanted to use mathematical induction to verify my suspicion.

I have assumed that $a_k<a_{k+1}$, I don't see how I can obtain $a_{k+1}<a_{k+2}$ because $\frac{1}{a_k}>\frac{1}{a_{k+1}}$

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    $\begingroup$ you can see that each time we add something positive so it is increasing! $\endgroup$
    – ahdahmani
    Sep 19, 2019 at 10:29
  • $\begingroup$ All terms are positive, so $a_n = a_{n-1} +1/a_{n-1} > a_{n-1}$. $\endgroup$
    – JavaMan
    Sep 19, 2019 at 10:30

5 Answers 5

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Base case:

$$1+\dfrac11>1\implies a_2>a_1.$$

Inductive step:

$$a_n=a_{n-1}+\frac1{a_{n-1}}>a_{n-1} \\\implies a_{n-1}>0 \\\implies a_n=a_{n-1}+\dfrac1{a_{n-1}}>0 \\\implies a_{n+1}=a_n+\frac1{a_n}>a_n.$$


Anyway, it is much simpler to establish $a_n>0$ ($1>0$ and $a_n>0\implies a_{n+1}=a_n+\dfrac1{a_n}>a_n>0)$ which is enough to justify the growth.

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The sequence is positive. Easy proof by induction.

Then $a_n - a_{n-1} = 1/a_{n-1} >0$ proving that the sequence is increasing.

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Hint, prove the following theorems in order:

Theorem 1. $a_n$ is positive.

Theorem 2. $a_n > a_{n-1}$

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The sequence is increasing.

Since $a_{n-1}>0, $ it follows that $\dfrac1{a_{n-1}}>0$, and therefore that $a_n=a_{n-1}+\dfrac1{a_{n-1}}>a_{n-1}$.

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Prove that $a_n > 0$ for all $n$.

Then use that $$ a_n = a_{n-1} + \frac{1}{a_{n-1}} > a_{n-1}, $$ since $1 / a_{n-1} > 0$.

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