Find an expression/upper bound for $$\sum_{r\le X}\frac{1}{m^r}{{k+r-1}\choose{r}}$$where $m,k$ are fixed.
Looking at the first few terms, we see that this is the same as $$\frac{1}{m}{{k}\choose{1}}+\frac{1}{m^2}{{k+1}\choose{2}}+\frac{1}{m^3}{{k+2}\choose{3}}+\cdots$$
I expect that the $1/m^a$ terms cause this sum to converge. I also wouldn't be surprised if this sum is actually known as a combinatorial result, so if anyone has a solution/reference that would be really helpful.
What I've tried so far is to brute force expand using the definition of $^nC_r$, as follows: $$\sum_{r\le X}\frac{1}{m^r}{{k+r-1}\choose{r}}=\sum_{r\le X}\frac{(k+r-1)!}{m^rr!(k-1)!},$$but this didn't seem to be a very fruitful approach.
