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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.

kotomord
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user6259845
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6 Answers6

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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.

Fred
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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.

Vincenzo Oliva
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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 $\,C\sim 0$.

Vlad
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  • Your argument is only valid if you know that $C \ne 0$. – Fred Mar 17 '17 at 07:57
  • @Fred Wouldn't contradiction still hold in case $C=0$? $$C = C + \frac{1}{C} \iff \frac{1}{C} = 0 \implies C\neq 0$$ – Vlad Mar 17 '17 at 08:00
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HINT: Show that $x_{n + \lceil 2 x_n \rceil} > 2 x_n$.

TMM
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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.

Elijah Moore
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As others have shown, $x_{n+1} \geq x_n > 0$ for all $n \in \mathbb{N}$. The important point that we will be using is the fact that each term is strictly positive (i.e., non-zero).

Suppose that $(x_n)$ is convergent. Let $L$ be its limit. $L$ is either zero or non-zero. We shall first show that $L \neq 0$. Consider the sequence $(x_{n+1}^2)$. If $L = 0$, notice that $\lim_{n\to\infty}x_{n+1}^2 = L^2 = 0$. For all $n$,

$$x_{n+1}^2 = \left(x_n + \frac{1}{x_n}\right)^2 = x_n^2 + \frac{1}{x_n^2} + 2 \geq 2$$

implying that $\lim_{n\to\infty}x_{n+1}^2 \geq 2$, which is a contradiction. Hence, $L \neq 0$.

Since $L \neq 0$,

$$ \lim_{n\to\infty}x_n = \lim_{n\to\infty}x_{n+1} = \lim_{n\to\infty}\left(x_n + \frac{1}{x_n}\right) = \lim_{n\to\infty}x_n + \frac{1}{\lim_{n\to\infty} x_n} $$

The last equality is true because $L \neq 0$ and $x_n \neq 0$. Hence,

$$ L = L + \frac{1}{L} \Leftrightarrow \frac{1}{L} = 0 $$

which is also a contradiction. Hence $(x_n)$ is divergent.

yihangho
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