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).

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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

Answer 3

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, $$ \sum\limits_{k = 1}^n {\left\{ n \atop k \right\}\frac{1}{{k + 1}}} = \sum\limits_{k = 0}^n {\binom{n}{k}\frac{{{\rm B}_{k + 1} }}{{k + 1}}B_{n - k} } . $$