Expression for $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}$? It is known that $\sum\limits_{k=0}^n \left\{ n \atop k\right\} k = \varpi(n+1) - \varpi(n)$. Any ideas for computing $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}$ ? ($\left\{ n \atop k\right\}$ denotes the Stirling numbers of the second kind and $\varpi(n)$ the $n$-th Bell number)
 A: We know 
$$B_n(x) = \sum_{k=0}^n \left\{ n \atop k\right\} x^k $$
where $B_n(x)$ is the Bell polynomial. Then
$$ \int_0^1 B_n(x) dx = \sum_{k=0}^n\frac1{k+1}\left\{ n \atop k\right\}$$
is what we are interested in computing. It's also known that
$$B_n(x) = e^{-x} \sum_{t=0}^{\infty} \frac{t^n x^t} {t!}$$
so we want the value of 
$$ \sum_{k=0}^n\frac1{k+1}\left\{ n \atop k\right\}= \sum_{t=0}^{\infty} \frac{t^n I(t)} {t!}, \hspace{1cm} I(t)= \int_0^1 e^{-x}x^t dx$$
But it can be shown (eg) that $I(t) \sim \frac{1}{e \,t}$, so, asymptotically 
$$ \sum_{k=0}^n\frac1{k+1}\left\{ n \atop k\right\} \sim e^{-1} \sum_{t=0}^{\infty} \frac{t^{n-1}}{t!} = B_{n-1}$$
which is the $n-1$-Bell number. Some values, taking the logarithm:
n     exact      approx
4    1.5114      1.6094
8    6.7348      6.7765
16  21.0308     21.0475
32  57.5872     57.5935

A: The error in leonbloy's approximation $\sum_{k=0}^n\frac1{k+1}\left\{ n \atop k\right\} =  B_{n-1} + E_n$ is exactly $$E_n = - \sum_{k=0}^{n-1}\left\{ n-1 \atop k\right\} \frac1{(k+1)(k+2)}.$$
Moreover, the asymptotic can be improved to $$\sum_{k=0}^n\frac1{k+1}\left\{ n \atop k\right\} \sim  B_{n-1} - B_{n-3},$$
(and probably further, for anyone who wants to continue the process below).

Theorem 4 of my paper "On Solutions to a General Combinatorial Recurrence" (Journal of Integer Sequences, 14 (9): Article 11.9.7, 2011), with $\left| {n \atop k} \right| = \left\{ n \atop k\right\}$ and $f(k,m) = \frac{1}{k+1}$ says that
$$\begin{align}\sum_{k=0}^n\frac1{k+1}\left\{ n \atop k\right\} &= \sum_{k=0}^{n-1}\left\{ n-1 \atop k\right\} (k+1) \frac1{k+1} - \sum_{k=0}^{n-1}\left\{ n-1 \atop k\right\} \frac1{(k+1)(k+2)} \\
&= B_{n-1} - \sum_{k=0}^{n-1}\left\{ n-1 \atop k\right\} \frac1{(k+1)(k+2)}. \end{align} $$

Applying Theorem 4 again, this time with $f(k,m) = \frac{1}{(k+1)(k+2)}$, yields (after some simplification)
$$\begin{align} &\sum_{k=0}^{n-1}\left\{ n-1 \atop k\right\} \frac1{(k+1)(k+2)} \\
&= \sum_{k=0}^{n-2}\left\{ n-2 \atop k\right\} \frac1{k+1} - \sum_{k=0}^{n-2}\left\{ n-2 \atop k\right\} \frac1{(k+1)(k+2)} - 2 \sum_{k=0}^{n-2}\left\{ n-2 \atop k\right\} \frac1{(k+1)(k+2)(k+3)}. \end{align}$$
The first term on the right-hand side dominates, and with leonbloy's approximation $$\sum_{k=0}^{n-2}\left\{ n-2 \atop k\right\} \frac1{k+1} \sim B_{n-3},$$
we get $$\sum_{k=0}^n\frac1{k+1}\left\{ n \atop k\right\} \sim  B_{n-1} - B_{n-3}.$$
By comparison with leonbloy's results (again, after taking logarithms)
n     exact      B_{n-1} - B_{n-3}     B_{n-1}
4    1.5114           1.3863            1.6094
8    6.7348           6.7154            6.7765
16  21.0308          21.0273           21.0475
32  57.5872          57.5866           57.5935

A: I will assume $n\ge 1$. Let $\widetilde{\rm B}_j$ denote the complementary Bell numbers. Using the notation from leonbloy's answer, we have
\begin{align*}
I(k) & = \int_0^1 {{\rm e}^{ - x} x^k {\rm d}x}  = \gamma (k + 1,1) = k\gamma (k,1) - {\rm e}^{ - 1} 
\\ & = {\rm e}^{ - 1} \sum\limits_{j = 1}^\infty  {\frac{1}{{(k + 1) \ldots (k + j)}}} 
\\ & \sim {\rm e}^{ - 1} \sum\limits_{j = 1}^\infty  {( - 1)^j \frac{{\sum\nolimits_{m = 1}^j {( - 1)^m \left\{ j \atop m\right\}} }}{{k^j }}}  = {\rm e}^{ - 1} \sum\limits_{j = 1}^\infty  {( - 1)^j \frac{{{\rm \widetilde B}_j }}{{k^j }}} 
\end{align*}
as $k\to +\infty$. Thus, if ${\rm B}_j$ denote the Bell numbers then
\begin{align*}
\sum\limits_{k = 1}^\infty  {\frac{1}{{k + 1}}\left\{ n \atop k\right\}} & \sim {\rm e}^{ - 1} \sum\limits_{k = 1}^\infty  {\frac{{k^n }}{{k!}}\sum\limits_{j = 1}^\infty  {( - 1)^j \frac{{{\rm \widetilde B}_j }}{{k^j }}} }  = \sum\limits_{j = 1}^\infty  {( - 1)^j {\rm \widetilde B}_j {\rm e}^{ - 1} \sum\limits_{k = 1}^\infty  {\frac{{k^{n - j} }}{{k!}}} } \\ & = \sum\limits_{j = 1}^\infty  {( - 1)^j {\rm \widetilde B}_j {\rm B}_{n - j} }  = {\rm B}_{n - 1}  - {\rm B}_{n - 3}  + {\rm B}_{n - 4}  + 2{\rm B}_{n - 5}  - 9{\rm B}_{n - 6}  +  \ldots ,
\end{align*}
as $n\to +\infty$.
An interesting expression in terms of the Bell numbers and the Bernoulli numbers $B_j$ may be obtained as follows. If
$$
C_n  = \sum\limits_{k = 1}^n {\left\{ n \atop k \right\}\frac{1}{{k + 1}}} 
$$
then
$$
\frac{1}{{n + 1}} = \sum\limits_{k = 1}^n s(n,k)C_k
$$
via properties of the Stirling transform. Then since
$$
f(x) = \sum\limits_{n = 1}^\infty  {\frac{1}{{n + 1}}\frac{{x^n }}{{n!}}}  = \sum\limits_{n = 1}^\infty  {\frac{{x^n }}{{(n + 1)!}}}  = \frac{1}{x}\sum\limits_{n = 2}^\infty  {\frac{{x^n }}{{n!}}}  = \frac{{{\rm e}^x  - x - 1}}{x},
$$
we have
\begin{align*}
\sum\limits_{n = 1}^\infty  {C_n \frac{{x^n }}{{n!}}} & = f({\rm e}^x  - 1) = \frac{x}{{{\rm e}^x  - 1}}\frac{{{\rm e}^{{\rm e}^x  - 1}  - 1}}{x} - 1 \\ &=  - 1 + \sum\limits_{n = 0}^\infty  {B_n \frac{{x^n }}{{n!}}} \sum\limits_{n = 0}^\infty  {\frac{{{\rm B}_{n + 1} }}{{n + 1}}\frac{{x^n }}{{n!}}} = \sum\limits_{n = 1}^\infty  {\left[ {\sum\limits_{k = 0}^n {\binom{n}{k}\frac{{{\rm B}_{k + 1} }}{{k + 1}}B_{n - k} } } \right]\frac{{x^n }}{{n!}}} .
\end{align*}
Consequently,
$$
\sum\limits_{k = 1}^n {\left\{ n \atop k \right\}\frac{1}{{k + 1}}} = \sum\limits_{k = 0}^n {\binom{n}{k}\frac{{{\rm B}_{k + 1} }}{{k + 1}}B_{n - k} } .
$$
