# Can the finite series $\sum_{k=1}^K \bigg(\frac{(m+k)!}{k!}\bigg)^2\bigg(\frac{1}{5}\bigg)^{2k}$ be approximated or bounded?

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?

• @metamorphy Ok, I see if I multiply my series by $1/(m!)^2$ I get the infinite series on the left hand side of your expression. But I don't see how that helps me as (i) I am dealing with a finite series and (ii) the series on the right hand side of your expression looks just as hard to approximate/bound as my original series. – csss Nov 8 at 16:31
• You can use my method to do the same thing by letting $$f(x,y)=\sum_{k=1}^\infty \left(\frac{(m+k)!}{k!}\right)^2x^ky^k.$$ – xpaul Nov 8 at 19:22
• @metamorphy To be more specific, $m > K \gg 1$. I would like to have a function (that doesn't feature a series) that accurately approximates the series, i.e. a function $f(m,K)$ such that $f(m,K) \approx \sum_{k=1}^\infty \frac{(m+k)!}{k!}\frac{1}{5^k}$. But if an approximation isn't possible an upper bound would be a good alternative. – csss Nov 8 at 19:29
• @xpaul But this question is about a finite series. You have specified a method for an infinite series, unless I'm missing something? – csss Nov 8 at 19:31
• Oh, that's a completely different story. – metamorphy Nov 8 at 20:51

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.
• I don't see how this will lead to the result as Wolfram Alpha says the result features the hypergeometric functions ${}_{2}F_1$ and ${}_{3}F_2$ so this is a much more complicated evaluation than my previous question. – csss Nov 8 at 20:03
• Wolfram Alpha gives the result in terms of these functions. Actually 2F1 arises for the infinite series, and then for the truncated series (if we ignore multiplying factors) WA gives 2F1 minus 3F2. So it seems the 3F2 term represents the truncated part of the infinite sum, i.e. $\sum_{k=K+1}^\infty$. – csss Nov 8 at 20:56