Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The task is to prove sequence convergence and find a limit. $x_0=0$ $x_1=1$ $x_{n+1}=\frac {x_n + n \cdot x_{n-1}} {n+1}$

I have computed some values of a sequence to build up some idea of the data: elements with even indexes converge from 0 to ~0.68, and elements with odd indexes converge from 1 to the same value.

It's obvious that the sequence isn't monotonic, hence I had to stick with a Cauchy theorem. But it led me nowhere: $|x_{n+p}-x_{n}| < | \frac {n \cdot x_{n-1}} {n + 1} + 2 \cdot \sum_{i=n}^{2n-2} x_i + { \frac {x_{2n-1}} {2n} } |$ (I got there under the assumption that $n = p$.)

Then I tried another move: $x_{n+1} - x_{n} = \frac {x_n + n \cdot x_{n-1}} {n+1} - x_n = $
$= \frac {-n \cdot ( x_n - x_{n-1})} {n+1}$
$x_{n} - x_{n-1} = \frac {(1-n) \cdot (x_{n-1} - x_{n-2})} {n}$

It looks like progress, but I still don't know how to go next.

share|cite|improve this question
This is the question from preliminary examination to Yadex's school of data analysis. Have you passed it? – DoctorMoisha Mar 12 '15 at 8:37
@DoctorMoisha, Yes. – wf34 Mar 12 '15 at 9:20
Congratulations! How was the third part - oral examination? Do they ask to prove something? I will be very grateful if you answer me even briefly. You can send me letter on and letter could be in Russian) Thank you! – DoctorMoisha Mar 12 '15 at 10:01
up vote 8 down vote accepted

You're on the right track. If you write $d_n=x_{n+1}-x_n$, you have $d_n=-\frac n{n+1}d_{n-1}$

So $d_n=(-\frac n{n+1})(-\frac {n-1}{n})(-\frac {n-2}{n-1})\ldots\frac12d_0=\frac{(-1)^n}{n+1}d_0=\frac{(-1)^n}{n+1}$

Then $x_n=x_0+\sum\limits_{k=0}^{n-1}d_k=\sum\limits_{k=0}^{n-1}d_k$

And $\lim\limits_{n\to\infty}x_n=\sum\limits_{n=0}^\infty d_n=\sum\limits_{n=0}^\infty \frac{(-1)^n}{n+1}$ which is a well known expansion of $\log 2$.

share|cite|improve this answer

as you have computed $x_{n+1}-x_n=-n(x_n-x_{n-1})/(n+1)$

we have $(n+1)(x_{n+1}-x_n)=-n(x_n-x_{n-1})$

then denote $g(n) = n(x_n-x_{n-1})$

we have $g(n+1) = -g(n)$

then $g(n) = (-1)^{n-1}g(1) = (-1)^{n-1}$

so $n(x_n-x_{n-1}) = (-1)^{n-1}$

$x_n-x_{n-1} = \frac{(-1)^{n-1}}{n}$

$x_n-x_m = \sum_{i=m+1}^{n}\frac{(-1)^{i-1}}{i}$

Now convergence of $x_n$ follows from convergence of $\sum{\frac{(-1)^n}{n}}$ and $x_n$ converges to $\ln{2}$

share|cite|improve this answer

The sequence converges to $\log 2$.

You can show inductively the following:

a. $x_{2n}$ is increasing,

b. $x_{2n-1}$ is decreasing,

c. $x_{n+1}-x_n=\dfrac{(-1)^n}{n+1}$.

d. $x_n=\displaystyle\sum_{k=0}^n \dfrac{(-1)^{k-1}}{k}\to\log 2$.

share|cite|improve this answer
@user2345215: Corrected! – Yiorgos S. Smyrlis Mar 4 '14 at 16:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.