Finding value of the limit: $ \lim_{t \to 0} \frac{\varphi(1+t) - \varphi(t)}{t}$ Let $$\varphi(x) = e^{-x} + \frac{x}{1!} \cdot e^{-2x} + \frac{3x^{2}}{2!}\cdot e^{-3x} + \frac{4^{2} \cdot x^{3}}{3!} \cdot e^{-4x} + \cdots$$
Then what is the value of: $$ \displaystyle\lim_{t \to 0} \frac{\varphi(1+t) - \varphi(t)}{t}$$
I am not getting any idea as to how to proceed for this problem. I tried summing up the expression but no avail. I also tried differentiating $\varphi(x)$ since the limit quantity which we require seems to involve derivative but again i couldn't find any pattern. Any ideas on how to solve this problem.
 A: The function $\varphi(x)$ can be written as
$$
\begin{eqnarray}
\varphi(x)
  &=& \frac{1}{x} \sum_{n=1}^{\infty} \frac{n^{n-1}}{n!}\left(xe^{-x}\right)^{n} \\
  &=& -\frac{1}{x} \sum_{n=1}^{\infty} \frac{(-n)^{n-1}}{n!}\left(-xe^{-x}\right)^{n} \\
  &=& -\frac{1}{x} W_0\left(-xe^{-x}\right),
\end{eqnarray}
$$
where $W_0$ is the Lambert W-function's main branch.  For sufficiently small $y$, the definition of the W-function ensures that $W_0(ye^y)=y$, so for small $t$,
$$
\begin{eqnarray}
\varphi(t)
  &=& -\frac{1}{t}W_0\left(-te^{-t}\right) \\
  &=& -\frac{1}{t}(-t) \\
  &=& 1.
\end{eqnarray}
$$
In fact this equality holds for all $0 \le t < 1$.  For values slightly larger than $1$, we use
$$
-(1+t)e^{-(1+t)} = \frac{-(1+t)}{1+t+\frac{1}{2}t^2+O(t^3)} = -1 + \frac{1}{2}t^2+O(t^3) = -(1-t)e^{-(1-t)},
$$
so
$$
\begin{eqnarray}
\varphi(1+t) &=& -\frac{1}{1+t}W_0(-(1+t)e^{-(1+t)}) \\ &\approx& -\frac{1}{1+t}W_0(-(1-t)e^{-(1-t)}) \\ &=& \frac{1-t}{1+t} \\ &=& 1 - 2t + O(t^2).
\end{eqnarray}
$$
The desired limit, then, is
$$
\lim_{t\rightarrow 0}\frac{\varphi(1+t) - \varphi(t)}{t} = \lim_{t\rightarrow 0}\frac{(1-2t) - 1}{t} = -2.
$$
A: Hmm, using Pari/GP, I just tried to find the coefficients of $ \varphi (t) $
polcoeffs(exp(-x)*sumalt(k=0,exp(-k*x)*x^k/k!*(k+1)^(k-1)*1.0))

(where the polcoeffs-function just retrieves the whole vector of coefficients up to the defined series-precision and I used x instead of t )
The result of this is an (arbitrary near) approximation to the formal powerseries $ \varphi(t) = 1.0 $ So it may be useful to analyze the approximation of that powerseries to the constant in order to evaluate  your limit-formula.    

[update] If one expands the powerseries for $e^{- k x}$ and collect coefficients at like powers of x then indeed all coefficients at a certain power of x sum to zero, thus vanish (at least heuristically up to $x^{64}$, I did not do final analysis) . Then all derivatives are also zero and even the L'Hospital-rule is of no use here. However, I don't know whether there could be some special reason that this expansion into powerseries might not be feasible/allowed because the result is an infinite sum of formal powerseries. [end update]    

[update 2]
$ \varphi(x)=e^{-x}+xe^{-2x}\frac1{1!} 
+x^2e^{-3x}\frac3{2!} 
+x^3e^{-4x}\frac{4^2}{3!} + \ldots
  $       
$\begin{array} {rrrr}
\varphi(x)= ( &  1 & -x & + \frac{x^2}{2!}&+ \frac{x^3}{3!} & + \ldots )\\
+ \frac{x}{1!}(& 1 & -2x & + \frac{2^2x^2}{2!}&+ \frac{2^3x^3}{3!} & + \ldots ) \\
+ \frac{3x^2}{2!}(& 1 & -3x & + \frac{3^2x^2}{2!}&+ \frac{3^3x^3}{3!} & + \ldots ) \\
+ \ldots \\
\end{array} $      
Collecting equal powers of x means to add coefficients along the antidiagonal so
$\begin{array} {rrrr}
\varphi(x)= 1 \\
- x (&\frac{1}{1!}& - \frac{1}{1!} ) \\
+ x^2 (&\frac{1}{2!}& - \frac{2}{1!}& + \frac{3^1}{2!}) \\
- x^3 (&\frac{1}{3!}& - \frac{2^2}{2!}& + \frac{3^2}{2!}& - \frac{4^2}{3!}) \\
+ \ldots \\
\end{array} $      
All the parentheses seem to form binomially weighted sums of like powers with alternating signs which are known to evaluate to zero, so the coefficients at each power of x should be zero and if this is so, the taylorseries is then $\varphi(x)=1 $
[end update 2]
