Hi,


Does anyone have an idea about an exact or approximate formulae for the following summation?
$$
\sum\_{j=1}^n \frac{j^k}{(j-1)!}
$$ 
where k is a positive integer (the denominator of the j^th term is of course $\Gamma(j)$).



Summation of an expression 
**REVISED ANSWER.**


In retrospect, deriving the approximation is quite easy. Indeed,
$$
\sum\limits\_{j = 1}^n {\frac{{j^k }}{{(j - 1)!}}} = e\sum\limits\_{j = 1}^n {e^{ - 1} \frac{{j^{k + 1} }}{{j!}}} \approx e\sum\limits\_{j = 0}^\infty {e^{ - 1} \frac{{j^{k + 1} }}{{j!}}} = eB\_{k + 1}.
$$
(For large $k$, you may consider [Asymptotic limit and bounds](http://en.wikipedia.org/wiki/Bell_number#Asymptotic_limit_and_bounds).)


**ORIGINAL ANSWER.**


Assume that $n$ is sufficiently large. Since
$$
\sum\limits\_{j = 1}^n {\frac{{j^k }}{{(j - 1)!}}} = \sum\limits\_{j = 0}^{n - 1} {\frac{{(j + 1)^k }}{{j!}}} , 
$$
the problem reduces to approximating $\sum\nolimits\_{j = 0}^{n - 1} {\frac{{j^m }}{{j!}}} $, for $0 \leq m \leq k$.
Now, $e^{ - 1} \sum\nolimits\_{j = 0}^\infty {\frac{{j^m }}{{j!}}} $ is the $m$-th moment of the Poisson distribution with mean $1$. For the latter, see, for example, [this](https://math.stackexchange.com/questions/6335/bell-numbers-and-moments-of-the-poisson-distribution).


**EDIT**: Specifically, the approximation (with the right-hand side being an upper bound) is
$$
\sum\limits\_{j = 1}^n {\frac{{j^k }}{{(j - 1)!}}} \approx e\sum\limits\_{m = 0}^k {{k \choose m}B\_m }, 
$$ 
where $B\_m$ is the $m$-th Bell number (a list is given [here](http://oeis.org/A000110)). Numerical results indicate that this approximation is very accurate, even for moderate values of $n$. For example, the absolute error for $n=15$, $k=3$ is $\approx 3.4 \times 10^{-9}$; for $n=20$, $k=5$ is $\approx 1.8 \times 10^{-12}$; for $n=20$, $k=7$ is $\approx 7.9 \times 10^{-10}$; for $n=23$, $k=9$ is $\approx 5.8 \times 10^{-11}$.


**EDIT**: If only the relative error is concerned, then the approximation is quite accurate even for relatively small $n$ values (of course, depending on $k$). For example, for $n=8$, $k=4$: $\approx 141.156$ compared to $\approx 141.351$; for $n=9$, $k=5$: $\approx 551.484$ compared to $\approx 551.811$; for $n=10$, $k=6$: $\approx 2383.359$ compared to $\approx 2383.933$.


**EDIT**: As Mike Spivey observed, using the identity
$$
\sum\limits\_{m = 0}^k {{k \choose m}B\_m } = B\_{k + 1}, 
$$
the above approximation can be simplified greatly to 
$$
\sum\limits\_{j = 1}^n {\frac{{j^k }}{{(j - 1)!}}} \approx e B\_{k+1}.
$$


