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Suppose we have a sequence defined by $a_{n+1}=\ln(a_{n}+2)$. We want to prove that for every $a_{0}>0$, the sequence converges to the same $g\in\mathbb{R}$.

This is where I got so far:

Let's define a function $f:[0,+\infty)\to\mathbb{R}$, $f(x)=\ln(x+2)$. First, let's note that $|f^{\prime}(x)|\leq 1/2$ for every $x\geq{0}$. From this, it follows that the inequality $|f(x)-f(y)|\leq\frac{1}{2}|x-y|$ holds for every $x,y\geq{0}$.

Now, let's return to our sequence. One can notice that the terms of the sequence become arbitrarily close to each other:

$$\left|a_{2}-a_{1}\right|=\left|f\left(a_{1}\right)-f\left(a_{0}\right)\right|\leq\frac{1}{2}\left|a_{1}-a_{0}\right|$$

$$\left|a_{3}-a_{2}\right|=\left|f\left(a_{2}\right)-f\left(a_{1}\right)\right|\leq\frac{1}{2}\left|a_{2}-a_{1}\right|=\frac{1}{2}\left|f\left(a_{1}\right)-f\left(a_{0}\right)\right|\leq\left(\frac{1}{2}\right)^{2}\left|a_{1}-a_{0}\right|$$

$$\vdots\qquad\qquad\vdots$$

$$\left|a_{n+1}-a_{n}\right|\leq\frac{1}{2}\left|a_{n}-a_{n-1}\right|\leq\left(\frac{1}{2}\right)^{n}\left|a_{1}-a_{0}\right|$$

Now, we have:

$$\left|a_{n+2}-a_{n}\right|\leq\left|a_{n+2}-a_{n+1}\right|+\left|a_{n+1}-a_{n}\right|$$

$$\leq\left(\frac{1}{2}\right)^{n}\left|a_{1}-a_{0}\right|+\left(\frac{1}{2}\right)^{n+1}\left|a_{1}-a_{0}\right|$$

$$=\left[\left(\frac{1}{2}\right)^{n}+\left(\frac{1}{2}\right)^{n+1}\right]\left|a_{1}-a_{0}\right|$$

$$\vdots\qquad\qquad\vdots$$

$$\left|a_{m}-a_{n}\right|\leq\left[\left(\frac{1}{2}\right)^{n}+\left(\frac{1}{2}\right)^{n+1}+...+\left(\frac{1}{2}\right)^{m-1}\right]\left|a_{1}-a_{0}\right|$$

$$=\left(\frac{1}{2}\right)^{n}\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+...+\left(\frac{1}{2}\right)^{m-1-n}\right]\left|a_{1}-a_{0}\right|$$

$$<\left(\frac{1}{2}\right)^{n}\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+...\right]\left|a_{1}-a_{0}\right|$$

$$=\left(\frac{1}{2}\right)^{n}\cdot\frac{1}{1-\frac{1}{2}}\cdot\left|a_{1}-a_{0}\right|$$

$$=\left(\frac{1}{2}\right)^{n-1}\left|a_{1}-a_{0}\right|$$

Thus, for every $\varepsilon>0$ we can find such $N$ that $\left(\frac{1}{2}\right)^{N-1}\left|a_{1}-a_{0}\right|<\varepsilon$. Such $N$ guarantees that for $m,n>N$ the inequality $|a_{m}-a_{n}|<\varepsilon$ also holds. This means that the sequence $a_{n}$ is Cauchy, which also means that it is convergent. This is where my dilemma starts: I believe that this just proves that for every $a_{0}>0$, the sequence $a_{n}$ is convergent to some limit, not necessarily the same one for every $a_{0}$. Am I wrong, and if not, then which path should I take right now?

Thanks in advance.

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You have a guarantee for the convergence. The you can take limit to the recurrence relation. So if you denote the limit as $\alpha$, then it must satisfy $\alpha = \log(\alpha+2)$. Now it remains to show that this equation has a unique zero. –  sos440 Sep 13 '12 at 17:32
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You don't really need that, @sos440, since $|f(x)-f(y)|<\alpha |x-y|$ where $0<\alpha<1$ implies the two have to converge to the same value, since $|a_{n}-b_{n}|<\alpha^n |a_0-b_0|$ –  Thomas Andrews Sep 13 '12 at 17:35
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@Thomas Andrews: I think I got your point... if we take two different points as $a_{0}$, $x$ and $y$, and if we have the sequence $a_{n}$starting in $a_{0}=x$ and $b_{n}$ starting in $b_{0}=y$, then $|a_{n}-b_{n}|\leq\alpha|f(a_{n-1})-f(b_{n-1})|\leq...\leq\alpha^{n}|a_{0}-b_{0}‌​|$. Right? –  Johnny Westerling Sep 13 '12 at 17:47
    
Yes, @JohnnyWesterling . –  Thomas Andrews Sep 13 '12 at 17:58
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2 Answers 2

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If $\lim_{n\to\infty}a_n=a$ then $a=\ln(a+2)$ because the logarithm is continuous. If $a, b$ are two solutions of $x=\ln(x+2)$, then the derivative of $x\mapsto x-\ln(x+2)$ must be zero for some point between $a$ and $b$.

With a bit more theory: Observe that $\ln(x+2)<x$ for $x>\sqrt 2$ (because $e^x=1+x+\frac12 x^2+\ldots>1+x+\frac12\cdot2=x+2$ for $x>\sqrt 2$). Therefore the function given by $f(x)=\ln(x+2)$ is a map of $K:=[0,\max\{a_0,\sqrt2\}]$ to itself. As you noted yourself, $|f'(x)|\le\frac12$ holds for all $x$, hence $|f(x_1)-f(x_2)|<\frac 12 |x_1-x_2|$ for all $x_1\ne x_2$, i.e. $f$ is a contraction. Then we use the Banach fixed-point_theorem: If $f:K\to K$ is a contraction on a compact set $K\subset\mathbb R$, then $f$ has a unique fixpoint $c$ (i.e. $f(c)=c$), and every sequence defined by $a_{n+1}=f(a_n)$ with $a_0\in K$ arbitrary converges to $c$.

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Hey, correct me if I'm wrong, but proving that the conditions formulated in Banach fixed-point theorem are met should spare me a lot of proving I did above... –  Johnny Westerling Sep 13 '12 at 18:07
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Let us go for the minimal technology: the function $f:x\mapsto\log(x+2)$ is (1.) increasing, (2.) such that $f(x)\lt x$ for every $x\lt g$, (3.) such that $f(x)\lt x$ for every $x\gt g$, for some $g\gt0$, (4.) continuous.

Thus $f(g)=g$. If $a_0=g$, $a_n=g$ for every $n$ hence $a_n\to g$. If $a_0\lt g$, $(a_n)_n$ is increasing and $a_n\lt g$ for every $n$ hence $(a_n)_n$ converges. The limit $\ell$ must be a fixed point of $f$ hence $\ell=g$, thus $a_n\to g$. Likewise if $a_0\gt g$, replacing increasing by decreasing and $a_n\lt g$ by $a_n\gt g$.

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