Determining when combinatorial sums are zero To keep things simple with a specific example, we ask: 
Prove that $\displaystyle\ a_n:=\frac{1}{n!}\sum_{k=0}^n \binom{n}{k} \frac{1}{k!} (-1)^{n-k}$ is zero if and only if $n=1$. (Or find a counter-example). 
In fact, determining when said sequence is positive and negative is even more challenging. We list the first few calculations, to show that it does not alternate in sign: 
$\{a_n\}_{n=0}^\infty=\{1, 0, -\frac{1}{4}, \frac{1}{9}, -\frac{5}{192}, \frac{7}{1800}, -\frac{37}{103680}, \frac{17}{2116800}, \frac{887}{232243200}, \ldots\}$
Posted a related question: https://mathoverflow.net/questions/211767/determing-signs-of-taylor-coefficients-in-entire-functions
 A: According to Maple, $a_n$ (with the revised definition) is $(-1)^n \text{LaguerreL}(n,1)/n!$.
I don't know asymptotics of this as $n \to \infty$,
but maybe something can be obtained from an integral representation of LaguerreL.
The plot is quite suggestive.

A: (Non-vanishing is established by multiplying by $n!^2$ and getting an expression which is 1 modulo $n$.)
Robert Israel's answer and the comments following it settle your problem completely, but here's a general strategy that often works:
Suppose you have a sum of the form $S(n) = \sum_{k=0}^{n} f(n,k)$, where $f$ is "hyper-geometric" in the sense that $f(n+1,k)/f(n,k), f(n,k+1)/f(n,k)$ are rational functions.


*

*Find the maximum of $|f(n,k)|$ as a function of $k$ ($n$ fixed). This is rather easy because $|f(n,k+1)|/|f(n,k)|$ is the absolute value of a rational function. In your specific problem, it is $\frac{n-k}{(k+1)^2}$, so $|f(n,k)|$ increases for $k \lesssim \sqrt{n}$ and then it decreases. The sign of the largest term is usually the resulting sign of the sum.

*Suppose the maximum is attained at $k=M(n)$. Focus on estimating the sum around $k=M(n)$ (the rest of the terms should be easier to bound, as we know they are smaller, the decay is exponential usually). Specifically, Stirling's approximation is usually enough for estimating $\sum_{k=-\Delta}^{\Delta} f(n,M(n)+k)$ - plugging Stirling and using some Taylor series estimates ($\Delta$ is chosen so that the Taylor series estimates are valid), we get a sum that can sometimes be evaluated even by comparing to a Riemann's integral. 


This is a rather elementary approach; Asymptotic methods such as steepest descent can yield much better results if you can express your sum as the coefficient of some analytic function. This can be done in your case as we have the relevant generating function: $e^{\frac{t}{1+t}-\ln(1+t)}$. See de Bruijn' "Asymptotic Methods in Analysis" (the example in section 4.7 is solved twice in the text - Laplace' method and Steepest descent, and it is reminiscent of your sum as it contains a sign-changing term).
The sign-changing property of your sum makes other nice methods obsolete. Hayman's method, for example, works in extracting the coefficients of $e^{P(t)}$ where $P$ is a function satisfying some positivity conditions ($\frac{t}{1+t}-\ln(1+t)$ is not admissible as its coefficients are sign-changing, but the method should succeed in approximating your sum if we remove the sign-changing term).
