Evaluate $\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$

Question

I'm interested in finding a nice closed form expression for the sum $\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$. I've tried using the Binomial Theorem to get \begin{align*} \sum_{k=1}^n\frac{1}{k}\binom{n}{k}x^k & =\int_0^1\frac{(1+x)^n-1}{x} \, dx\\ &=\int_1^2 (1+u+\cdots+u^{n-1}) \, du \end{align*} using the substitution $u=1+x$ but I can't quite to simplify this integral either. I have also not been able to come up with a combinatorial approach, which may not exist since the summation and its terms are in general not integers. Any help in evaluating this sum would be appreciated, thanks!

Answer 1

In the question, the problem is in "nice closed form expression" since
$$\sum_{k=1}^n\frac{1}{k}\binom{n}{k}x^k=n x \, _3F_2(1,1,1-n;2,2;-x)$$ where appears an hypergoemetric function.

So, let us forget the $x$ and compute for a few values of $n$ $$S_n=\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$$ $$\left\{1,\frac{5}{2},\frac{29}{6},\frac{103}{12},\frac{887}{60},\frac{1517}{60},\frac{18239}{420}\right\}$$ which are $$\left\{1,\frac{5}{2},\frac{29}{6},\frac{206}{24},\frac{1774}{120},\frac{18204}{720},\frac{218868}{5040}\right\}$$ The denominators are clearly $n!$ and the numerators corresponds to sequence A103213 in OEIS.

As you will see in the link is that, for large $n$ $$S_n\approx \frac{2^{n+1}} n$$ It is also given that $$S_n=-H_n-\Re(B_2(n+1,0))$$ where appear the harmonic number and the real part of the incomplete beta function.

Update

Concerning the asymptotic behavior, it seems that it could be slightly improved using $$S_n\approx 2^{n+1} n^{\frac{1}{4 n}-1}$$

Answer 2

Here's a way to get the same result, and another form as well, without integrals: $$ \begin{align} f(n) &=\sum_{k=1}^n\frac1k\binom{n}{k}\\ &=\sum_{k=1}^n\frac1k\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]\\ &=f(n-1)+\sum_{k=1}^n\frac1k\binom{n-1}{k-1}\\ &=f(n-1)+\frac1n\sum_{k=1}^n\binom{n}{k}\\ &=f(n-1)+\frac{2^n-1}n\\ &=\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^n\frac{2^k-1}k}\\ &=\sum_{k=1}^n\frac1k\sum_{j=0}^{k-1}2^j\\ &=\sum_{j=0}^{n-1}2^j\sum_{k=j+1}^n\frac1k\\ &=\bbox[5px,border:2px solid #C0A000]{\sum_{j=0}^{n-1}2^j(H_n-H_j)} \end{align} $$ However, I don't see a simple closed form.

Answer 3

Here's the best I can (currently?) do.

Continuing on from where you left off, namely $$S = \sum_{k = 1}^n \frac{1}{k} \binom{n}{k} = \int_1^2 (1 + u + \cdots + u^{n - 1}) \, du, \quad (*)$$ summing the finite geometric series we have \begin{align*} S = \sum_{k = 0}^{n- 1} \int_1^2 u^k \, du= \sum_{k = 0}^{n - 1} \frac{2^{k + 1}}{k + 1} - \sum_{k = 0}^{n - 1} \frac{1}{k + 1}= 2 \sum_{k - 0}^{n - 1} \frac{2^k}{k + 1} - H_n. \end{align*} Here $H_n$ are the Harmonic numbers.

The sum appearing on the right can be written in the terms of the Lerch transcendent $\Phi(z,s,a)$ as follows.

From the identity $$\Phi (z,s,a) = z^n \Phi(z,s,a+n) + \sum_{k = 0}^{n - 1} \frac{z^k}{(k + a)^s},$$ setting $a = 1, s = 1$, and $z = 2$, after rearranging, we have $$\sum_{k = 0}^{n - 1} \frac{2^k}{k + 1} = \Phi(2,1,1) - 2^n \Phi(2,1,n+1).$$ Thus $$\sum_{k = 0}^n \frac{1}{k} \binom{n}{k} = 2\Phi(2,1,1) - 2^{n + 1} \Phi(2,1,n+1) - H_n.$$ Note each Lerch transcendent appearing in the above expression is complex whose imaginary parts end up canceling out.

($*$) As the number of terms appearing in the integral is finite, the expression is actually in closed formed already. However, I think what you require is an expression that can be written more "compactly" in terms of known (special) functions.

Answer 4

$f(n)=∑ ^n_{k=1}\frac{1}{k} C^n_k $

This is equal to term by term multiplication of following series:

$∑^n_{k=1} C^n_k =∑^n_{k=0} C^n_k-1=(1+1)^n-1=2^n-1$

$∑^n_{k=1}\frac{1}{k}= 1+1/2+1/3 + . . . 1/n =H_n $

⇒ $f(n)=∑ ^n_{k=1}\frac{1}{k} C^n_k =(2^n-1)H_n=∑^n_{k=1}\frac{2^k-1}{k}$

Note that $(2^n-1)$represent$(∑ ^n_{k=1} C^n_k)$ and $(H_n)$ represent$(∑ ^n_{k=1}\frac{1}{k})$ that is $(∑ ^n_{k=1}\frac{1}{k} C^n_k)=∑ ^n_{k=1}\frac{1}{k}\times∑ ^n_{k=1} C^n_k$(term by term multiplication).For example for n=5:

$f(5)=∑ ^5_{k=1}\frac{1}{k} C^5_k$ is not$(2^5-1)(1+1/2+1/3+1/4+1/5)=\frac{4247}{60}$, it is calculated as:

$f(5)=∑ ^5_{k=1}\frac{1}{k} C^5_k=\frac{2^1-1}{1}+\frac{2^2-1}{2}+\frac{2^3-1}{3}+\frac{2^4-1}{4}+\frac{2^5-1}{5}=\frac{887}{60}$