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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 :)

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2 Answers 2

up vote 2 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

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

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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
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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 $0$.

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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
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