Question

\[\text{ A sequence } a_n \text{ is defined recursively by the equation } \\ a_0=a_1=1 \\ n(n-1)a_n=(n-1)(n-2)a_{n-1}-(n-3)a_{n-2} \\ \text{ Find the sum of the series } \\ \sum_{n=0}^{\infty} a_n \]

Answer 1

This should be a better link. I hope.

http://www.purplemath.com/modules/nextnumb3.htm

Answer 2

lol you are trying your best with helpful links
but this sequence and sum we have to find is a little more difficult than what is usually posted on purple math (or what I have seen anyways)

Answer 3

If you know how to do it, Please teach me o-o

Answer 4

might be another telescoping series

Answer 5

or not lol

Answer 6

\[a_n=\frac{(n-2)a_{n-1}}{n}-\frac{(n-3)a_{n-2}}{n(n-1)} \]

Answer 7

i think splitting the second fraction might help

Answer 8

\[a_n=\frac{(n-2)a_{n-1}}{n}-\frac{(n-3)a_{n-2}}{n-1} + \frac{(n-3)a_{n-2}}{n} \]

Answer 9

\[a_{n-1}=\frac{(n-3)a_{n-2}}{n-1}-\frac{(n-4)a_{n-3}}{n-2}+\frac{(n-4)a_{n-3}}{n-1}\]

Answer 10

so if we add that term with the previous term we get...

Answer 11

oh that one fraction cancels

Answer 12

I still don't understand any of this...

Answer 13

we need four fractions in a_n itself to use telescoping.. wid odd fractions it wont work

Answer 14

\[a_0+a_1+a_2+a_3+a_4+a_5 + \cdots \\ =1+1+\frac{1}{2}+\frac{1}{6}+(\frac{1}{12}-\frac{1}{24})+(\frac{1}{20}-\frac{1}{40}-\frac{1}{60})+ \cdots\]
I think I found those first few five terms correctly

but this doesn't really help

also this is another even question no answer in back

Answer 15

unfortunately I didn't give myself enough time to look at this
i have to go
i hope you don't have too much fun without me :p

Answer 16

These are the terms I got (from a_0 to a_5):
\[a_n = 1, 1, \frac{1}{2}, \frac{1}{6}, \frac{1}{24}, \frac{1}{120}, \ldots\]
Which can be simplified as
\[a_n = \frac{1}{n!}\]
Hopefully that can help.

Answer 17

looks nice xD