In my previous question I asked about methods of evaluating the following infinite series: $$ \sum_{k=1}^\infty \frac{(m+k)!}{k!}\frac{1}{5^k}. $$
I now have a somewhat related question. This time I am interested in the following finite series: $$ \sum_{k=1}^K \bigg(\frac{(m+k)!}{k!}\bigg)^2\bigg(\frac{1}{5}\bigg)^{2k}. $$ I am interested in methods of finding a good approximation or tight bound for this series? Note that $m$ is an integer and $m > K \gg 1$, but I don't know if this is any use. So is is possible to get an accurate approximation/bound for this series?
Let $$ f(x,y)=\sum_{k=1}^K \left(\frac{(m+k)!}{k!}\right)^2x^ky^k.$$ Then integrating $M$ times w.r.t $x$ and $y$, respectively gives \begin{eqnarray} F(x,y)&:=&\int\cdots\int\int\cdots\int f(x,y)dx\cdots dxdy\cdots dy\\ &=&\sum_{k=1}^K (xy)^{m+k}+\sum_{j,k=0}^mC_jD_kx^jy^k\\ &=&\frac{(xy)^{m+1}(1-(xy)^K)}{1-xy}+\sum_{j,k=0}^{m-1}C_jD_kx^jy^k\\ &=&\frac{1}{1-xy}\bigg[\sum_{k=0}^{m+1}\binom{m+1}{k}(xy-1)^{k}-\sum_{k=0}^{m+K+1}\binom{K}{k}(xy-1)^{k}\bigg]+\sum_{j,k=0}^{m-1}C_jD_kx^jy^k\\ &=&-\bigg[\sum_{k=1}^{m+1}\binom{m+1}{k}(xy-1)^{k-1}-\sum_{k=1}^{m+K+1}\binom{K}{k}(xy-1)^{k-1}\bigg]+\sum_{j,k=0}^{m-1}C_jD_kx^jy^k\\ &=&-\bigg[(xy)^m-\sum_{k=m+1}^{m+K+1}\binom{K}{k}(xy-1)^{k-1}\bigg]+P(x,y) \end{eqnarray} where $P(x,y)$ is a polynomial of $x$ and $y$ of degree $m-1$. Repeating the method in Methods of evaluating $ \sum_{k=1}^\infty \frac{(m+k)!}{k!}\frac{1}{5^k}$? I used, you can do the rest.