The recurrence relation $$a_n=u\cdot a_{n-1}+vn$$ has the special solution $$a_n=\frac{v}{1-u}n-\frac{uv}{(u-1)^2}$$ for $u\ne 1$
The recurrence relation $$a_n=a_{n-1}+vn$$ has the special solution $$a_n=\frac{vn(n+1)}{2}$$
The ansatz $a_n=rn+s$ is successfull in the case $u\ne 1$, but in the case $u=1$, we need a polynomial of degree $2$ in $n$
How can I understand this phenomen and how can I find such solutions in general ?
Is it true that the general solution of $a_n=u\cdot a_{n-1}+vn$ (which is $a_n=c\cdot u^n+\frac{v}{1-u}n-\frac{uv}{(u-1)^2}$) tends to the general solution of $a_n=a_{n-1}+vn$ (which is $c+\frac{vn(n+1)}{2}$) , if $u$ tends to $1$, or is this a fallacy ?