# Did I compute the limit of of the sequence $x_{n+1}=\frac{x_n}{x_n+1}, x_o=1$ properly?

I need to study the limit behavior of $x_{n+1}=\frac{x_n}{x_n+1}, x_o=1$ and if the limit exists, compute the limit. I observed the first few terms and it seemed that the sequence was decreasing so I decided to show that the sequence was monotonic:

Need to show $x_{n+1}\leq x_n$ which is equivalent to showing $x_{n+1}-x_n\leq0$ $$x_{n+1}-x_n=\frac{x_n}{x_n+1}-x_n=\frac{x_n-x_n(x_n+1)}{x_n+1}=\frac{-(x_n)^2}{x_n+1}=-\frac{(x_n)^2}{x_n+1}\leq0$$ for all n if $x_n\geq0$ which can be proved by induction: $$x_o=1\geq0$$$$x_1=1/2\geq0$$ $x_n=>x_{n+1}$ $$x_{n+1}=\frac{x_n}{x_n+1}>\frac{1}{x_n+1}>0$$ because $x_n>0$. (I don't know if that was the proper way to do the induction. Any confirmation?)

Since it was proved that $x_n\geq0$, $x_{n+1}-x_n=-\frac{(x_n)^2}{x_n+1}\leq0$, thus the sequence is monotone and decreasing. The sequence is also bounded:

Since the sequence is decreasing it is bounded above by 1, and because $x_n\geq0$ the sequence id bounded below by 0.

The boundedness and monotonicity of the sequence implies that a limit exists:

Let $\lim x_n=x$. Because $\lim x_n=\lim x_{n+1}$, $$x=\frac{x}{x+1}<=>x(x+1)=x<=>x+1=1=>x=0$$ So $0$ is the limit.

I'm not sure if there are problem in the work that I did and any help would be greatly appreciated.

I don't know if I should start a new question for this, so I've included it here anyways: Since one was able to tell that the $\lim x_n=\lim \frac{1}{1+n}$ for the above sequence, should one try to do the same thing with the sequence $x_{n+1}= \frac{(x_n)^2}{x_n+1}$, $x_o=1$?

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You could also show that your sequence is $x_n=\frac{1}{n}$. –  Baby Dragon Mar 16 '13 at 20:45
Also note that your initial c0ndition, $x_0$ does not matter as long as $x_0=-1$ is avoided. –  Baby Dragon Mar 16 '13 at 20:52
The $x_o=1$ was given to me in the question. Why doesn't it matter? –  user66807 Mar 16 '13 at 20:58
@user66807 because the sequence is $x_n = \frac{1}{n+1}$. This sequence converges, irrespective of the choice of $x_0$. All you need is: $x_0 > 0$. (Some definitions of naturals numbers are the positive integers anyhow) . –  Rustyn Mar 16 '13 at 21:01
@Rustyn Yazdanpour Oh I see, makes sense now. Thanks. –  user66807 Mar 16 '13 at 21:04

There’s a small error in your proof that $x_n\ge 0$ for all $n$: it’s not true that

$$\frac{x_n}{x_n+1}>\frac1{x_n+1}\;,$$

since in fact it turns out that $x_n\le 1$ for all $n$. However, given the induction hypothesis that $x_n\ge 0$, you certainly have\

$$x_{n+1}=\frac{x_n}{x_n+1}\ge 0\;,$$

which is all you need here. Otherwise it looks fine.

Note that if you calculate the first few values, you find that $x_1=\frac12$, $x_2=\frac13$, and $x_3=\frac14$, suggesting that in general $x_n=\frac1{n+1}$. An alternative approach would be to show by induction that this is true for all $n\ge 0$; this is not at all difficult.

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Thanks for pointing out my mistake. Even though I worked out the first several terms of the sequence I failed to realize that $x_n=\frac1{n+1}$. Thanks for pointing that out, too. –  user66807 Mar 16 '13 at 21:03
@user66807: You’re welcome. –  Brian M. Scott Mar 16 '13 at 21:04

Denote $y_n=1/x_n$ and you can find $y_n$.

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$$x_n = \frac{1}{n+1}$$
Base case: $x_0 = \frac{1}{0+1}$
$$\text{LHS} = x_0 = 1 = \frac{1}{0+1} = \text{RHS}$$ Assume that for $k$: $$x_k = \frac{1}{k+1}$$ Note that $x_{k+1} = \dfrac{\dfrac{1}{k+1}}{\dfrac{1}{k+1} +1 } = \dfrac{1}{k+1}\cdot \dfrac{k+1}{k+2} = \dfrac{1}{k+2}$

This closes the induction.
Thus $$\lim_{n\to\infty} x_n = \lim_{n\to\infty} \frac{1}{n+1} = 0$$

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Thanks for the induction, it helps. –  user66807 Mar 16 '13 at 21:17
@user66807 No problemo –  Rustyn Mar 16 '13 at 21:24