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Let $a_0 > 0$

$a_{n+1} = \ln(a_n + 1)$

Prove that $\displaystyle\lim_{n\to\infty} n\cdot a_n = 2$

Przemysław Scherwentke
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user192276
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    Did you try to prove than $a_n \rightarrow 0$ first? You could use limited expansion then, that would help you I think. Hint: $a_n$ >0 , $a_n$ decreases.. – mvggz Nov 14 '14 at 09:01

2 Answers2

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If you're not used to this type of question, here's a process that often works:

1) Prove that here your sequence is decreasing, and bounded:

$\ln(1+x) \leq x \implies a_{n+1} \leq a_n $ , ie $(a_n)$ decreases.

Now you can prove by a trivial induction that $(a_n) \geq 0$.

$(a_n)$ is lower-bounded, decreasing, hence it converges $\rightarrow L$.

2) Evaluate L:

You can take the limit on both side of the recursive relation : $L = \ln(L+1) $

If you study the function $h: x \rightarrow x - \ln(1+x)$ , you'll see that the only possibility to get h(x) = 0 is : x=0. That proves that : L=0.

3) Pushing it further: $\ln(1+u) = u - \frac{u^2}{2} + o(u^2)$ ; when $u \rightarrow 0$

You get : $ a_{n+1} = a_n - \frac{a_n^2}{2} + o(a_n^2) = a_n(1 - \frac{a_n}{2} + o(a_n)) $

Let:$ v_n = a_n^{-1}$

=> $ v_{n+1} = v_n(1- \frac{a_n}{2} + o(a_n))^{-1} $

$(1-u)^{-1} = 1 + u +o(u)$ ; when $ u \rightarrow 0 $, thus giving:

$ v_{n+1} = v_n(1 +\frac{a_n}{2} + o(a_n)) = v_n + \frac{1}{2} + o(1) $

=> $ v_{n+1} -v_n = \frac{1}{2} + o(1) $

Here you can say that partial sums of both part are equivalent, since $\sum \frac{1}{2}$ diverges to $+\infty$.

You get: $ v_{n} - v_0 $ ~ $v_n$ ~ $\frac{1}{2}\sum_{k=0}^{n-1} 1 = \frac{n}{2}$

Finally you get : $ a_n = v_n^{-1} $ ~$ \frac{2}{n}$ ; that is: $ na_n \rightarrow 2$

mvggz
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  • The two lines beginning with "If you study the function..." are incomprehensible to me: first, what does "a function being null" mean? Second, why would we be interested of forced to consider $;h;$ a null function? Third, how this proves $;L=0;$ ?? – Timbuc Nov 14 '14 at 15:34
  • I'm sorry I misexpressed myself, but on the other hand given the previous lines you could have deduced that I was just looking for solutions to the equation: h(x) = 0 , since the limit verifies : h(L) = 0 Now if you want me to detail this : $h'(x) = 1 - \frac{1}{1+x}$ >0 whenever x>0. h(0) =0 , then h(x) > 0 if x > 0 . If you're still not convinced, you might note that f(x) = x is the tangent to h at x=0 , and that h is concave, so it is under its tangents. – mvggz Nov 14 '14 at 15:48
  • Thanks @mvggz, I did know that. I simply didn't understand your terminology. There is also a weird double minus sign in the second line of (3). – Timbuc Nov 14 '14 at 15:49
  • @Timbuc I misused the word "null", sorry.. But I think the rest is good right? Thanks for the remark, I've edited – mvggz Nov 14 '14 at 15:51
  • I think so, @mgvvz, typos apart. +1 – Timbuc Nov 14 '14 at 15:52
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For limit calculation propose another variant that is a combination of Lemma Stolz-Cesaro and rule of L'Hospital:

$$\lim_{n\rightarrow\infty}na_n=\lim_{n\rightarrow\infty}\frac{n}{\frac{1}{a_n}}=\lim_{n\rightarrow\infty}\frac{n+1-n}{\frac{1}{a_{n+1}}-\frac{1}{a_n}}=\lim_{n\rightarrow\infty}\frac{a_n\cdot a_{n+1}}{a_n-a_{n+1}}=\lim_{n\rightarrow\infty}\frac{a_n\cdot \ln (1+a_n)}{a_n-\ln (1+a_n)}=\lim_{x\rightarrow0}\frac{x\cdot \ln (1+x)}{x-\ln (1+x)}=\lim_{x\rightarrow0}\frac{ln(1+x)}{x}\cdot\frac{x^2}{x-\ln (1+x)}=1\cdot\lim_{x\rightarrow0}\frac{2x}{1-\frac{1}{1+x}}=2.$$

medicu
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