# Determine if sequence described by $a_1=1$, $a_n=a_{n-1}+\frac{1}{a_{n-1}}$ is increasing

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

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

## 5 Answers

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.

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.

Hint, prove the following theorems in order:

Theorem 1. $$a_n$$ is positive.

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

The sequence is increasing.

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

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 $$1 / a_{n-1} > 0$$.