cjquines0 wrote: The sequence ${a_0, a_1, a_2, ...}$ of real numbers satisfies the recursive relation $$n(n+1)a_{n+1}+(n-2)a_{n-1} = n(n-1)a_n$$for every positive integer $n$, where $a_0 = a_1 = 1$. Calculate the sum $$\frac{a_0}{a_1} + \frac{a_1}{a_2} + ... + \frac{a_{2008}}{a_{2009}}$$. Immediate induction shows that $a_k=\frac 1{k!}$ and so required sum is $\sum_{k=0}^{2008}\frac{(k+1)!}{k!}=\boxed{1005\times 2009}$

Solution: For $n=1$,we see that, $$1.2.a_2-a_0=0 \implies a_2=\frac{1}{2}=\frac{1}{2!}$$Now, suppose that, $$a_k=\frac{1}{k!} \forall k \in \mathbb{N}$$is valid for all numbers $\leq k$. Then we now put $n=k$, then, $$k(k+1)a_{k+1}+(k-2)\frac{1}{(k-1)!}=k(k-1).\frac{1}{k!}$$$$\implies k(k+1)a_{k+1}=\frac{1}{(k-1)!} \implies a_{k+1}=\frac{1}{(k+1)!}$$So, now we have, $$\sum_{i=0}^{2008} \frac{a_i}{a_{i+1}}=\sum_{i=1}^{2008} (i+1) =1005*2009=2019045$$And, we are done. $\blacksquare$

But how e can guess the ${{a}_{n}}=\frac{1}{n!}$?

TuZo wrote: But how e can guess the ${{a}_{n}}=\frac{1}{n!}$? I tried $n=1,2,3$, they came like $a_2=\frac{1}{2},a_3=\frac{1}{6},a_4=\frac{1}{24}$, so it was obvious.

You can also notice that $a_n=\frac{1}{n!}$ works since it fits the recursion: \[\frac{n(n+1)}{(n+1)!}+\frac{n-2}{(n-1)!}-\frac{n(n-1)}{n!}=\frac{n(n+1)+n(n+1)(n-2)-n(n+1)(n-1)}{(n+1)!}=\frac{(\cancel{n^2}+n)+(\cancel{n^3}-\cancel{n^2}-2n)-(\cancel{n^3}-n)}{(n+1)!}=\frac{0}{(n+1)!}=0\] Then it is clear since $\frac{1}{0!}=\frac{1}{1!}=1$. This is similar to the induction idea above!