Expression for $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}$?

Question

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)

Answer 1

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

Answer 2

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