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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 :)

2 Answers 2

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Well

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

and $x_1 = 1$ so

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

  • 1
    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 $x_n = x_{(n-1)+1} = \frac{1}{2^{n-1}(n-1)!}$.2012-10-06
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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 $0$.

  • 2
    Even if you added $1$ to all the $x(n)$, the sequence would still be decreasing and positive. But its limit wouldn't be $0$.2012-10-06