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.



Alternative proofs sought after for a certain identity
$\DeclareMathOperator\lead{leader} \DeclareMathOperator\prob{prob}$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.


