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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 ?

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Peter
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    The solution of the recursion $$a_n=u\cdot a_{n-1}+vn$$ starting from $a_0=c$ is $$a_n=\left(c+\frac{uv}{(u-1)^2}\right)\cdot u^n+\frac{v}{1-u}n-\frac{uv}{(u-1)^2}$$ This is also $$a_n=cu^n+\frac{uv}{(u-1)^2}\cdot (u^n-1)+\frac{v}{1-u}n$$ or, equivalently, $$a_n=cu^n+v\frac{f_n(u)-f_n(1)}{u-1}$$ with $$f_n(u)=u^{n}+\cdots+u^2+u$$ Thus, when $u\to1$, $$a_n\to c+vf'n(1)$$ where $$f'_n(1)=n+\cdots+2+1=\frac{(n+1)n}2$$ that is, $$\lim{u\to1}a_n=c+v\frac{(n+1)n}2$$ and the limit is indeed the solution of $a_n=a_{n-1}+vn$ with initial condition $a_0=c$. – Did May 14 '17 at 08:51
  • @Did Thank you very much for this informative comment! – Peter May 14 '17 at 08:56

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