One competition in Croatia

[rsoldo]

Is a series $(a_n)$ and

$$n\cdot a_n-a_{n-1}-(n-1)\cdot a_{n-2}=\frac{2n-3}{(n-1)(n-2)}$$

for $n\geq 3$ convergent? If yes, find his limit.

My hint:Let ${{a}_{n}}-{{a}_{n-1}}={{x}_{n}},\text{ then }n{{x}_{n}}={{y}_{n}}$, and then ${{y}_{n}}-\frac{1}{n-1}={{z}_{n}}$, and you will get ${{z}_{k}}+{{z}_{k-1}}=0\Leftrightarrow \frac{{{z}_{k}}}{{{z}_{k-1}}}=-1/\prod\limits_{k=2}^{n}{\Rightarrow {{z}_{n}}}={{(-1)}^{n-1}}$

[kotenok gav]

Ну так вы же дали решение.

[rsoldo]

Sorry, sorry, I forgot one 'detail' $a_1=1$ and $a_2=\frac{3}{2}$.

[rsoldo]

Solution:

$$n\cdot a_n-a_{n-1}-(n-1)\cdot a_{n-2}=\frac{2n-3}{(n-1)(n-2)}$$

$$n\cdot a_n-n\cdot a_{n-1}+(n-1)\cdot a_{n-1}-(n-1)\cdot a_{n-2}=\frac{1}{n-1}+\frac{1}{n-2}$$

$$n\cdot (a_n-a_{n-1})+(n-1)\cdot(a_{n-1}-a_{n-2})=\frac{1}{n-1}+\frac{1}{n-2}$$

We introduce substitution: $x_{n-1}=a_n-a_{n-1}$

$$n\cdot x_{n-1}+(n-1)\cdot x_{n-2}=\frac{1}{n-1}+\frac{1}{n-2}.$$

We substitute again: $y_n=(n+1)\cdot x_n-\frac{1}{n}$, and we have $y_{n-1}+y_{n-2}=0$ and $y_1=0$.
Now we have $y_n=0$ for all $n\in \mathbb{N}$, and $x_n=\frac{1}{n\cdot (n-1)}$.
Now:

$$a_n=a_{n-1}+\frac{1}{n\cdot (n-1)}=...=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=2-\frac{1}{n}$$

Therefore, the series is monotonous and bounded from above:

$$L= \lim_{n\to\infty} a_n=2.$$