More sequence help

**relyt** · April 12th 2009, 06:34 PM

Show that sequence $r_n$ defined by the formula:

$r_n = \frac {(-1)^n} {n!}$

satisfies the recurrence

$r_k = \frac {-r_{k -1}}{k}$

This is another one where I really don't know where to start. Any help would be appreciated.

THanks

**K A D C Dilshan** · April 12th 2009, 06:53 PM

Originally Posted by relyt

Show that sequence $r_n$ defined by the formula:

$r_n = \frac {(-1)^n} {n!}$

satisfies the recurrence

$r_k = \frac {-r_{k -1}}{k}$

This is another one where I really don't know where to start. Any help would be appreciated.

THanks

This is a simple Qn .Isn`t it ?
$r_n = \frac {(-1)^n} {n!}$

$r_{n-1} = \frac {(-1)^{n-1}} {{n-1}!}$

$\frac {r_n}{r_{n-1}} = \frac { \frac {(-1)^n} {n!}}{\frac {(-1)^{n-1}} {(n-1)!}}$

then
$\frac {r_n}{r_{n-1}} = \frac{-1}{n}$

so it satisfies the above recurrence formula .

**xalk** · April 12th 2009, 07:17 PM

Originally Posted by relyt

Show that sequence $r_n$ defined by the formula:

$r_n = \frac {(-1)^n} {n!}$

satisfies the recurrence

$r_k = \frac {-r_{k -1}}{k}$

This is another one where I really don't know where to start. Any help would be appreciated.

THanks

.

We have:

$r_{n-1} =\frac{\frac{(-1)^n}{(-1)}}{(n-1)!}=-\frac{(-1)^n}{(n-1)!}$ .................................................. ...........................1

Now multiply (1) by 1/n and we get:

$-\frac{r_{n-1}}{n} = \frac{(-1)^n}{n(n-1)!} = r_{n}$

**relyt** · April 12th 2009, 07:41 PM

Thanks to both of you for answering, but which is correct? I seem to get lost about half way though

**K A D C Dilshan** · April 12th 2009, 07:45 PM

Originally Posted by relyt

Thanks to both of you for answering, but which is correct? I seem to get lost about half way though

both answers are correct bro ,
but I think my answer is more descriptive .
Have you not get it yet ?
simply starts from $r_n$ and $r_{n-1}$ do the proof .

**relyt** · April 12th 2009, 08:27 PM

Ok, i think I understand...I'll try some more

THanks

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