<s>[INST] Alternative proofs sought after for a certain identity 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.
 [/INST] 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.
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