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

I'm trying to find a way to solve the following difference equation, but I have exhausted all the resources at my disposal so now I come here for guidance. The equation is the following:

$$x_1 = 1,\quad x_{n+1}={x_n \over 2n},\ n>1.$$

Is there a general method for solving equations like these?

Thanks for reading :)

share|cite|improve this question
up vote 3 down vote accepted


$$x_{n+1} = \frac{1}{2n}x_n = \frac{1}{2n}\frac{1}{2(n-1)}x_{n-1} = \dots = \frac{1}{2n}\frac{1}{2(n-1)}\dots\frac{1}{2(2)}\frac{1}{2(1)}x_1,$$

and $x_1 = 1$ so

$$x_{n+1} = \frac{1}{2n}\frac{1}{2(n-1)}\dots\frac{1}{2(2)}\frac{1}{2(1)} = \frac{1}{2^n(n\ . (n-1)\dots 2\ . 1)} = \frac{1}{2^nn!}.$$

share|cite|improve this answer
What you've done make total sense but the solution manual says $$ x_{n} = {1 \over (n-1)!2^{n-1}} $$ – L1meta Oct 6 '12 at 17:11
They are equivalent. The solution guide gives the formula for $x_n$ whereas I gave the formula for $x_{n+1}$. Using what I've done, we have $x_n = x_{(n-1)+1} = \frac{1}{2^{n-1}(n-1)!}$. – Michael Albanese Oct 6 '12 at 17:20

You could easily demonstrate by induction that:

$$x_{n+1}=\frac { x(0) }{ { 2 }^{ n }n! }$$

Now, observing that $x(0)=1$, you can state that $x_n$ is decrescent and positive, and its limit is $0$.

share|cite|improve this answer
Even if you added $1$ to all the $x(n)$, the sequence would still be decreasing and positive. But its limit wouldn't be $0$. – TonyK Oct 6 '12 at 16:06

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.