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given is the following sequence:

$a_1 > 0$

$a_n = \frac{1}{1+a_n}$

I succeeded in finding a (possible?) limit by guessing that the sequence is limited by a; then the sequence $a_n$ converges to a; however, the sequence $a_n = \frac{1}{1+a_n}$ also converges to a. This means: $a = \frac{1}{1+a}$. Solving this equation gives the 'possible' limit.

However, is this the end? Shouldn't I first show THAT the sequence actually IS convergent? If so: HOW should I do this?

I can show:

$0 < a_n < 1$

I cannot show:

$a_n$ is monotonous.

Vazrael
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2 Answers2

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If $a=\frac1{1+a}$, then $$\tag1a_{n+1}-a=\frac1{1+a_n}-\frac1{1+a}=\frac{a-a_n}{(1+a_n)(1+a)} $$ Now if $a>0$ and $a_n>0$, this implies that $$ |a_{n+1}-a|\le \frac1{1+a}\cdot |a_n-a|=q\cdot |a_n-a|$$ with $0<q=\frac1{1+a}<1$. Then by induction $$|a_n-a|\le q^{n-1}\cdot |a_0-a|$$ i.e. $a_n\to a$.

Remark: From $(1)$ we see that $a_n-a$ changes sign at each step, hence the sequence $(a_n)_{n\in\mathbb N}$ is indeed not monotonic. But subsequences $(a_{2n})$ and $(a_{2n+1})$ are.

Hagen von Eitzen
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  • Regarding the remark: Indeed. Starting with $a_0 = 0$, you get the convergents of the continued fraction of the golden ratio, which alternate above and below the limit. – Aryabhata May 06 '13 at 17:01
  • Ok, thank you very much. I have to take a closer look to this, but I'm sure I'll get it soon :) – Vazrael May 07 '13 at 10:10
1

From $0 \lt a_n \lt 1$ you have proven that the sequence has at least one accumulation point in the interval. What could go wrong is that the sequence could bound around the neighborhoods of several points. To show that it is a limit, write $a_n=a+e_n$, so $e_n$ is the error of term $n$. Then $e_{n+1}=a_{n+1}-a=\frac 1{1+a+e_n}-\frac 1{1+a}=\frac {-e_n}{(1+a+e_n)(1+a)}$. As long as $e_n$ is not too large and negative (I leave it to you to argue this) the series will converge to zero, and we have a limit.

Ross Millikan
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