Alternative proofs sought after for a certain identity

Question

Here is an identity for which I outlined two different arguments. I'm collecting further alternative proofs, so

QUESTION. can you provide another verification for the problem below?

Problem. Prove that $$\sum_{k=1}^n\binom{n}k\frac1k=\sum_{k=1}^n\frac{2^k-1}k.$$ Proof 1. (Induction). The case $n=1$ is evident. Assume the identity holds for $n-1$. Then, \begin{align*} \sum_{k=1}^{n+1}\binom{n+1}k\frac1k-\sum_{k=1}^n\binom{n}k\frac1k &=\frac1{n+1}+\sum_{k=1}^n\left[\binom{n+1}k-\binom{n}k\right]\frac1k \\ &=\frac1{n+1}+\sum_{k=1}^n\binom{n}{k-1}\frac1k \\ &=\frac1{n+1}+\frac1{n+1}\sum_{k=1}^n\binom{n+1}k \\ &=\frac1{n+1}\sum_{k=1}^{n+1}\binom{n+1}k=\frac{2^{n+1}-1}{n+1}. \end{align*} It follows, by induction assumption, that $$\sum_{k=1}^{n+1}\binom{n+1}k\frac1k=\sum_{k=1}^n\binom{n}k\frac1k+\frac{2^{n+1}-1}{n+1}=\sum_{k=1}^n\frac{2^k-1}k+\frac{2^{n+1}-1}{n+1} =\sum_{k=1}^{n+1}\frac{2^k-1}k.$$ The proof is complete.

Proof 2. (Generating functions) Start with $\sum_{k=1}^n\binom{n}kx^{k-1}=\frac{(x+1)^n-1}x$ and integrate both sides: the left-hand side gives $\sum_{k=1}^n\binom{n}k\frac1k$. For the right-hand side, let $f_n=\int_0^1\frac{(x+1)^n-1}x\,dx$ and denote the generating function $G(q)=\sum_{n\geq0}f_nq^n$ so that \begin{align*} G(q)&=\sum_{n\geq0}\int_0^1\frac{(x+1)^n-1}x\,dx\,q^n =\int_0^1\sum_{n\geq0}\frac{(x+1)^nq^n-q^n}x\,dx \\ &=\int_0^1\frac1x\left[\frac1{1-(x+1)q}-\frac1{1-q}\right]dx=\frac{q}{1-q}\int_0^1\frac{dx}{1-(x+1)q} \\ &=\frac{q}{1-q}\left[\frac{\log(1-(1+x)q)}{-q}\right]_0^1=\frac{\log(1-q)-\log(1-2q)}{1-q} \\ &=\frac1{1-q}\left[-\sum_{m=1}^{\infty}\frac1mq^m+\sum_{m=1}^{\infty}\frac{2^m}mq^m\right]=\frac1{1-q}\sum_{m=1}^{\infty}\frac{2^m-1}m\,q^m \\ &=\sum_{n\geq1}\sum_{k=1}^n\frac{2^k-1}k\,q^n. \end{align*} Extracting coefficients we get $f_n=\sum_{k=1}^n\frac{2^k-1}k$ and hence the argument is complete.

Answer 1

$\DeclareMathOperator\lead{leader} \DeclareMathOperator\prob{prob}$Answering a follow-up question by Per Alexandersson. Here is the $q$-version obtained by a suitable modification of the probabilistic proof of the OP identity.

We consider the linear space $X:=\mathbb{F}_q^n$ over a finite field $\mathbb{F}_q$. For $x=(x_1,\ldots,x_n)\in X\setminus {0}$ denote $\lead(x)=\max(i:x_i\ne 0)$, for a subspace $L\subset X$, $m:=\dim L>0$, denote $\lead(L)=\max_{x\in L} \lead(x)$. It follows from Gauss elimination that $L$ contains a basis $f_1,\ldots,f_m$, such that $\lead(f_1)<\lead(f_2)<\ldots <\lead(f_m)=\lead(L)$. Thus $L$ contains exactly $q^m-q^{m-1}$ elements $x$ for which $\lead(x)=\lead(L)$, and $q^{m-1}$ 1-dimensional subspaces $R$ for which $\lead(R)=\lead(L)$.

Choose a random subspace $L$ of $X:=\mathbb{F}_q^n$ with probability of $k$-dimensional subspace proportional to $q^{k\choose 2} y^k$ ($k=1,2,\ldots,n$). The sum of these weights is $(1+y)(1+qy)\ldots (1+q^{n-1}y)-1:=\theta_n$ (that's $q$-binomial theorem).

Then choose a random 1-dimensional subspace $R\subset L$ uniformly. Consider the following probability: $\kappa:=\prob(\lead(R)=\lead(L))$. On one hand, $$ \kappa=\sum_{k=1}^n \prob(\dim L=k)\prob(\lead R=\lead L|\dim L=k)\\ =\theta_n^{-1}\sum_{k=1}^n q^{k\choose 2}y^k{n\choose k}_q\cdot \frac{q^{k-1}}{[k]_q}, $$ where $[k]_q=1+q+\ldots+q^{k-1}$ is the number of 1-dimensional subspaces of a $k$-dimensional space over $\mathbb{F}_q$.

On the other hand, denoting by $X_k$ the $k$-dimensional subspace of $x\in X$ for which $x_{k+1}=\ldots=x_n=0$, we get $$ \kappa=\sum_{k=1}^n \prob(L\subset X_k\& \lead(R)=k)=\sum_{k=1}^n \prob(L\subset X_k)\cdot \prob(\lead(R)=k|L\subset X_k)\\= \theta_n^{-1}\sum_{k=1}^n \theta_k\cdot \frac{q^{k-1}}{[k]_q}. $$

Thus the identity (we multiply both expressions for $\kappa$ by $q\cdot \theta_n$) $$ \sum_{k=1}^n q^{k+1\choose 2}y^k{n\choose k}_q\cdot \frac{1}{[k]_q}=\sum_{k=1}^n \frac{q^{k}((1+y)(1+qy)\ldots (1+q^{k-1}y)-1)}{[k]_q}, $$ for $q=y=1$ we get the initial identity.

Answer 2

$\DeclareMathOperator\prob{prob}$Alapan Das' clever argument may be rephrased on the probabilistic language. Write $[m]=\{1,2,\dotsc,m\}$. Choose a random non-empty subset $A\subset [n]$ (all $2^n-1$ possible outcomes having equal probabilities). Then choose a random element $\xi\in A$ uniformly. Denote $p=\prob (\xi=\max(A))$. On one hand, denoting $j=\lvert A\rvert$ we get $$ p=\sum_{j=1}^n \prob(\xi=\max(A)\mathrel||A|=j)\cdot \prob(|A|=j)=\sum_{j=1}^n\frac1j \cdot\frac{{n\choose j}}{2^n-1} $$

On the other hand, $$ p=\sum_{k=1}^n \prob(\xi=k \, \&\,A\subset [k])= \sum_{k=1}^n \prob(\xi=k\mathrel|A\subset [k])\cdot \prob(A\subset [k])\\= \sum_{k=1}^n \frac1k\cdot \frac{2^k-1}{2^n-1}. $$

Answer 3

\begin{align*} \sum_{k=1}^{n} \frac{2^k-1}{k} &=\sum_{k=1}^{n} \frac{1}{k}\left(\sum_{j=1}^{k} \binom{k}{j}\right) \\ &=\sum_{j=1}^{n} \sum_{k=j}^{n} \binom{k}{j}\frac{1}{k} \\ &=\sum_{j=1}^{n} \frac{1}{j}\left(\sum_{k=j}^{n} \binom{k-1}{j-1}\right) \\ &=\sum_{j=1}^{n} \frac{1}{j} \binom{n}{j}. \end{align*}

Answer 4

$$\sum_{k=1}^n\binom nk\frac1k=\sum_{k=1}^n\binom nk\int_0^1 dt\,t^{k-1}= \int_0^1 dt\,\sum_{k=1}^n\binom nk t^{k-1}=\int_0^1 dt\,\frac{(1+t)^n-1}t.$$ $$\sum_{k=1}^n\frac{2^k-1}k=\sum_{k=1}^n\int_0^1 dt\,(1+t)^{k-1}= \int_0^1 dt\,\sum_{k=1}^n (1+t)^{k-1}=\int_0^1 dt\,\frac{(1+t)^n-1}t.$$

Answer 5

One can also use the binomial transform.

(If $A(z)=\sum_{i\geq 0} a_i z^i$ is a (formal) power series, the (formal) power series $B(z):=\frac{1}{1-z} A(\frac{z}{1-z})$ has coefficients $[z^n] B(z)=\sum_{j=0}^n {n \choose j} a_j$).

We have $\log(\frac{1}{1-z})=\sum_{k\geq 1} \frac{z^k}{k}$.

Thus \begin{align*} \sum_{k=1}^n {n \choose k}\frac{1}{k}&=[z^n] \frac{1}{1-z}\,\log\big(\frac{1}{1-\frac{z}{1-z}}\big)\\ &=[z^n] \frac{1}{1-z}\,\log\big(\frac{1-z}{1-2z}\big)\\ &=[z^n] \frac{1}{1-z}\,\Big(\log\big(\frac{1}{1-2z}\big)-\log\big(\frac{1}{1-z}\big)\Big)\\ &=\sum_{k=1}^n\frac{2^k}{k} -\sum_{k=1}^n \frac{1}{k}\end{align*}

Answer 6

Here's a sketch of a proof of a generalization:

\begin{multline} \quad \sum_{k=1}^n\binom nk \frac{t^k}{k+a}\\ =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}. \quad \tag {$*$} \end{multline}

(This is a generalization of Terry Tao's generalization, which is the case $a=0$.)

We start with the identity $$\sum_{k=0}^n \binom nk \frac{t^k}{k+a} = \frac {1}{a\binom{a+n}{n}}\sum_{k=0}^n \binom{a+k-1}{k} (1+t)^k.$$ This is a special case of a well-known linear transformation for the hypergeometric series, the case $b=a+1$ of $${}_2F_1(-n,a; b\mid -t) =\frac{(b-a)_n}{(b)_n}\,_2F_1(-n,a; 1-n-b+a\mid 1+t),$$ where $(u)_n = u(u+1)\cdots (u+n-1)$, which can be proved easily in several ways.

Since $\frac{1}{a}\binom{a+k-1}{k} = \frac {1}{k}\binom{a+k-1}{k-1}$ for $k\ge 1$, we have $$\sum_{k=1}^n\binom nk \frac{t^k}{k+a} =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}+C$$ where $C$ is a constant (as a polynomial in $t$). But $C=0$ since each summand has no constant term in $t$, and $(*)$ follows.