Summation to n of Power of k over k

Theorem

 * $\displaystyle \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \displaystyle \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\left({x - 1}\right)^k} k$

Proof
The proof proceeds by induction over $n$.

For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \displaystyle \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\left({x - 1}\right)^k} k$

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_{k \mathop = 1}^r \dfrac {x^k} k = H_r + \displaystyle \sum_{k \mathop = 1}^r \dbinom r k \dfrac {\left({x - 1}\right)^k} k$

from which it is to be shown that:
 * $\displaystyle \sum_{k \mathop = 1}^{r + 1} \dfrac {x^k} k = H_{r + 1} + \displaystyle \sum_{k \mathop = 1}^{r + 1} \dbinom {r + 1} k \dfrac {\left({x - 1}\right)^k} k$

Induction Step
This is the induction step:

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 1}: \displaystyle \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \displaystyle \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\left({x - 1}\right)^k} k$