Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge?

Question

Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.

I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a >0$. Then...

$x_2 = 2 + \frac{1}{2} = 2.5$

$x_3 = 2.5 + \frac{1}{2.5} = 2.9$

$x_4 = 2.9 + \frac{1}{2.9} = 3.2448$

$x_5 = 3.2448 + \frac{1}{3.2448} = 3.5530$

$x_6 = 3.5530 + \frac{1}{3.5530} = 3.8344$

$x_7 = 3.8344 + \frac{1}{3.8344} = 4.0952$

$x_8 = 4.0952 + \frac{1}{4.0952} = 4.3394$

$x_9 = 4.3394 + \frac{1}{4.3394} = 4.5698$

$x_{10} = 4.5698 + \frac{1}{4.5698} = 4.7887$

but others have said it converges so I'm confused on whether it converges or diverges? Can someone please explain.

Answer 1

Assume the sequence converges i.e. $\displaystyle\lim_{n\to \infty} x_n = C< \infty.$ Then, as $n\to\infty$ we have $\,x_{n}\approx x_{n+1}$. Taking limit of $x_{n+1}=x_n+\frac{1}{x_n}$ we get

$$ \lim_{n\to\infty} x_{n}= \lim_{n\to\infty} \left(x_n+\frac{1}{x_n} \right)\implies C = C + \frac{1}{C} \iff \frac{1}{C} = 0 \quad \text{ – } \text{ contradiction} $$

The last expression results in contradiction even if $\,C\sim 0$.

Answer 2

It is easy to see (by induction) that $x_{n+1}>x_n \ge a >0$ for all $n$.

We assume that $(x_n)$ is convergent and that $l$ is the limmit. Then $l \ge a>0$.

From $x_{n+1}= x_n + (1/x_n)$ we get $l= l+ 1/l$, which is impossible. Hence the sequence is divergent.

Answer 3

HINT: Show that $x_{n + \lceil 2 x_n \rceil} > 2 x_n$.

Answer 4

We have $$x_{N+1}-x_1=\sum_{n=1}^N x_{n+1}-x_n =\sum_{n=1}^N \frac1{x_n}$$ so $x_n$ converges if and only if the rightmost series converges.

But the series converges only if $x_n$ diverges to $\pm\infty $, impossible: both the sequence and the series must diverge.

Answer 5

We can see the series $x_{n+1}$ is monotone increasing as $n\rightarrow \infty$ using the Monotone Convergence Theorem:

Monotone Increasing $\Rightarrow x_{n+1}>x_{n}, \forall n$

$\Rightarrow x_n+\frac{1}{x_n}>x_n$

$\Rightarrow \frac{1}{x_n} > 0$

Which is true since $x_1=a$ and $a>0$.

So now we know we have an increasing sequence. Lets assume some limit $x$ exists, which implies $x_{n+1}=x_n$, so we solve to 'find' our limit:

$x=x+\frac{1}{x}\Rightarrow x^2=x^2+1$, leading to a contradiction.

Therefore the limit does not exist and the sequence diverges.