$\sum\limits_{k=1}^n \frac{x^k}{k}=H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}$ Prove:
$$\sum\limits_{k=1}^n \frac{x^k}{k}=H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}$$
where $H_n=\sum\limits_{k=1}^n\frac1k$
For the $x^k$ i tried  $x^k=((x-1)+1)^k$ and decompose it into binomial expansion but I got dual sum at the one side one sum at the other side
For the second method:
$$\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}-\sum\limits_{k=1}^n \frac{x^k}{k}=H_n$$
and try to show
$$\binom{n}{k}(x-1)^k-x^k=1$$
Either the problem is wrong or I cannot do it.
I think the identity does not hold. Using
$$ x^k=[1+(x-1)]^k=\sum_{i=0}^k\binom{k}{i}(1-x)^i=1+\sum_{i=1}^k\binom{k}{i}(1-x)^i $$
one has
\begin{eqnarray}
&amp;&amp;\sum\limits_{k=1}^n \frac{x^k}{k}\\
&amp;=&amp;\sum\limits_{k=1}^n\frac{1}{k}\bigg(1+\sum_{i=1}^k\binom{k}{i}(1-x)^i\bigg)\\
&amp;=&amp;H_n+\sum\limits_{k=1}^n\sum_{i=1}^k \binom{k}{i} \frac{(x-1)^i}{k}\\
&amp;\neq&amp;H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}.
\end{eqnarray}
This identity does hold true
express both the RHS and LHS as integrals
then interchange the integrals and summations.
Hope this helps