1. More sequence help

Show that sequence $\displaystyle r_n$ defined by the formula:

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

satisfies the recurrence

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

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

THanks

Originally Posted by relyt
Show that sequence $\displaystyle r_n$ defined by the formula:

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

satisfies the recurrence

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

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

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

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

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

so it satisfies the above recurrence formula .

3. Originally Posted by relyt
Show that sequence $\displaystyle r_n$ defined by the formula:

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

satisfies the recurrence

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

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

THanks
.

We have:

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

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

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

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

5. 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 $\displaystyle r_n$ and $\displaystyle r_{n-1}$ do the proof .

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

THanks