Find $a_{n}$ if $(a_{n})$ is a sequence such that $a_{1} := 1$ and $\frac{1}{a_{n+1}} = \frac{2}{a_{n}} + 3$ for $n \geq 2$?

Question

This problem is weird. By the initial condition $a_{1} = 1$ we have $a_{2} = \frac{1}{5}$ and so on. But is there really a pattern for $a_{n}$?

I guess this problem is that kind of problems that require some luck to see the solution?

Hint: Note that $$a_{n+1}^{-1}+3=2(a_n^{-1}+3)=\cdots=2^{n}(a_1^{-1}+3)=2^{n+2}.$$ — Did, Mar 16 '15 at 12:07
@Did: That is it! Thanks. For others' reference convenience, would you mind augmenting this laconic comment to be an answer? — Yes, Mar 16 '15 at 12:14
Here is another idea: expand my comment into a full answer and post it here. Yes, this is quite encouraged by the official guidelines of the site. Furthermore, after a decent delay so that everybody has the time to criticize/validate your post, you could even accept it. — Did, Mar 16 '15 at 12:19

lab bhattacharjee · Accepted Answer · 2015-03-16T11:46:00.870

3

HINT:

Let $1/a_n=b_n+cn+d\implies b_1=\cdots$

So, we have $b_{n+1}+c(n+1)+d=2\left(b_n+cn+d\right)+3$

$\iff b_{n+1}=2b_n+cn+d+3$

Set $d+3=0,c=0$

edited Mar 16 '15 at 11:46

answered Mar 16 '15 at 11:37

lab bhattacharjee

Appreciate. But would you please explain a little more why the choice? – Yes Mar 16 '15 at 11:39
@Chou, First of all, the choice inversion is obvious, right? Then If $1/a_{n+1}-2/a_n=a+bn+cn^2,b_m=1/a_m+A+Bm+Cm^2+Dm^3$ one greater than the highest power of difference – lab bhattacharjee Mar 16 '15 at 11:42
Sorry, I still did not get your idea. :) – Yes Mar 16 '15 at 11:50
Since $c = 0$ and $d = -3,$ we have $1/a_{n} = b_{n} - 3,$ and then? – Yes Mar 16 '15 at 12:02
@lab FYI, I cannot make head or tail of your first comment. – Did Mar 16 '15 at 12:05
@Chou, $$b_{n+1}=2b_n=\cdots=2^rb_{n-r}$$ for integers $n,r$ right? – lab bhattacharjee Mar 16 '15 at 12:06
@Did, I meant if $a_{n+1}=A\cdot a_n+O(n^m)$ Set $b_n=a_n+O(n^{m+1})$ – lab bhattacharjee Mar 16 '15 at 12:08
Yes, probably, and? – Did Mar 16 '15 at 12:11
@Did, Then we can easily define iterative definition of $b_n$ from the recursive ones. Ex: http://math.stackexchange.com/questions/807306/solving-a-recursion-relation-a-n1-3a-n4a-n-12/807343#807343 and http://math.stackexchange.com/questions/975064/this-recurrence-relation-will-evaluate-to/975073#975073 etc. – lab bhattacharjee Mar 16 '15 at 12:12
Sure, we could, but this is not the problem at all. The problem is whether your rendition of this ultra-classical approach is understandable to this OP or not. Personally, although I know all this, I find difficult to decode what you mean to say, so... – Did Mar 16 '15 at 12:17

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