Method for solution of the recurrence $a_{n+1} = ka_n + n^{-c}$ ($c > 0$)?

Question

As stated in the title, I am wondering if there is a general method for the solution of the recurrence

$$a_{n+1} = ka_n + n^{-c}$$

where $c > 0$.

I've tried a couple of the standard approaches, without success.

Answer 1

\begin{align} a_{n+1} &= k(ka_{n-1} + (n-1)^{-c}) + n^{-c}\\ &=k^2a_{n-1} + k(n-1)^{-c}+n^{-c}\\ &=\dots\\ &=k^{n+1}a_0+\sum_{i=0}^n k^i(n-i)^{-c} \end{align}

Answer 2

just a hint try telescoping,

$$a_{n+1}=ka_n+1/n^c$$ $$ka_n=k^2a_{n-1}+k/(n-1)^c $$

$$k^2a_{n-1}=k^3a_{n-2}+k^2/(n-2)^c $$

$$k^{n-1}a_2=k^na_1+k^{n-1}/1$$

by summing

$$a_{n+1}=k^n\Bigl(a_1+\sum_{i=1}^n\frac {1}{i^ck^i}\Bigr)$$