Find closed form for $a_n=2\frac{n-1}{n}a_{n-1}-2\frac{n-2}{n}a_{n-2}$ for all $n \ge 3$

Question

Find a closed form for:

$a_n=2\frac{n-1}{n}a_{n-1}-2\frac{n-2}{n}a_{n-2}$ for all $n \ge 3$

Initial values: $a_1=3, a_2=1$

I'm struggling a bit with the algebra as the solution contains complex number. Any help would be appreciated.

Answer 1

Remark that it satisfies $$ na_n=2\left(\left(n-1\right)a_{n-1}-\left(n-2\right)a_{n-2}\right) $$ Then I recommend you study the sequence $\left(b_n\right)_{n \in \mathbb{N}}$ defined by $$ b_n=na_n $$ which satisfies $$ b_n=2\left(b_{n-1}-b_{n-2}\right) $$ Do you know how to deal with this kind of sequence ?

EDIT :

The sequence is what we call linear induction sequence of order $2$ that I prefer to rewrite for $n \in \mathbb{N}^{*}$ $$ b_{n+2}-2b_{n+1}+2b_n=0 $$ You search for geometrical sequences that satisfy it, it leads you to solve the equation $$ r^2-2r+2=0 $$ The discriminant equals to $\Delta=4-8=-4<0$. Here comes the complex solution that you were talking about ; the two solutions are $$ r_1=\frac{2+2i}{2}=1+i \ \text{ and } \ r_2=\frac{2-2i}{2}=1-i $$ Then those two geometrics are solutions, and it is a two-dimensional space so we got the solution $$ b_n=\alpha\left(1+i\right)^{n-1}+\beta \left(1-i\right)^{n-1} $$ So using $b_n=na_n$ it comes that your sequence that looks not easy to compute is in fact verifying for all $n \in \mathbb{N}$ $$ a_n=\frac{\alpha\left(1+i\right)^{n-1}+\beta \left(1-i\right)^{n-1}}{n} $$ Then $a_1=3$ and $a_2=1$ so $$ \alpha+\beta=3 \ \text{ and } \ 2=\alpha\left(1+i\right)+\beta\left(1-i\right) $$ You inject the first in the second $$ 2=\alpha\left(1+i\right)+\left(3-\alpha\right)\left(1-i\right) \Leftrightarrow 2i\alpha=3i-1 \Leftrightarrow \alpha=\frac{3}{2}+\frac{i}{2} $$ And $$\beta=3-\alpha=\frac{3}{2}-\frac{i}{2}$$

Hence this dreadful sequence is given for $n \in \mathbb{N}^{*}$ by

$$ a_n=\left(\frac{3}{2}+\frac{i}{2}\right)\frac{\left(1+i\right)^{n-1}}{n}+\left(\frac{3}{2}-\frac{i}{2}\right)\frac{\left(1-i\right)^{n-1}}{n} $$

Hope it helped you, if you have any question on how to proceed, dont hesitate.

Answer 2

Set $u_n = na_n $ then your relation becomes

$$u_n = 2u_{n-1}-2u_{n-2}$$

Whose characteritic equation is $$x^2-2x+2=0 \implies x=\frac{2\pm 2i}{2}$$

And hence $$ u_n =A\left(\frac{2-2i}{2}\right)^{n-1} +B\left(\frac{2+2i}{2}\right)^{n-1}$$

that is $$\color{red}{a_n =\frac{A}{n}\left(\frac{2-2i}{2}\right)^{n-1} +\frac{B}{n}\left(\frac{2+2i}{2}\right)^{n-1}}$$ Now can you fine A and B?

Answer 3

My approach is similar to the others but somewhat easier to facilitate. We start off, of course, by taking $f_n=na_n$ so that we have the sequence

$$f_n=2f_{n-1}-2f_{n-2}$$

This is a generalized Fibonacci form, i.e., $f_n=a f_{n-1}+bf_{n-2}$, for which I have previously shown here that a general solution is given by

$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{af_0}{2} \frac{\alpha^n+\beta^n}{\alpha+\beta}=\frac{(f_1-f_0\beta)\alpha^n-(f_1-f_0\alpha)\beta^n}{\alpha-\beta}\quad n=0,1,2,\cdots$$

or

$$f_n=\frac{(f_2-f_2\beta)\alpha^{n-1}-(f_2-f_1\alpha)\beta^{n-1}}{\alpha-\beta}\quad n=1,2,3,\cdots$$

where $\alpha,\beta$ are the characteristic roots given by

$$\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$$

Thus, substituting $f_1=a_1$ and $f_2=2a_2$, along with $\alpha,\beta=1\pm i$ into the above, and then $a_n=f_n/n$, I find that

$$a_n=\frac{(-1+3i)(1+i)^{n-1}-(-1-3i)(1-i)^{n-1}}{2ni}$$

I have verified this solution numerically.