$ \sum_{m=1}^{n} \sum_{k=1}^{m} (-1)^{k+1} H_k ~ ^{n}C_{k} = H_n $ How to show that $$\sum_{m=1}^{n}\sum_{k=1}^{m} (-1)^{k+1} H_k  ~ {m \choose k} = H_n.$$
 I substituted $ H_k $ with $ \int_{0}^{1} \frac{1-x^{(k-1)}}{1-x} dx$ in the summation, but I got stuck in integration of the summation.
 A: Let $$S_n=\sum_{m=1}^{n}\sum_{k=1}^{m} (-1)^{k+1} H_k  ~ {m \choose k}$$
 Using the representation of the harmonic numbers $$  H_k = \int_{0}^{1} \frac{x^{k}-1}{x-1} dx. ~~~(1) $$ Notice  the power of $x$ is $k$ and not $k-1$ in (1),  we get, $$ S_n = \sum_{m=1}^{n} \int_{0}^{1} \sum_{k=1}^{m} (-1)^{k+1} ~ \frac{x^{k}-1}{x-1} ~ {m \choose k} ~ dx = \sum_{m=1}^{n} \int_{0}^{1} \frac{1}{x-1} \sum_{k=1}^{m} \left[- (-x)^k {m \choose k}+(-1)^{k} {m \choose k} \right]dx 
$$ $$ \Rightarrow S_n = \sum_{m=1}^{n} \int_{0}^{1} \frac{1}{x-1} [1 -(1-x)^m -  1 ] ~dx = \sum_{m=1}^{n} \int_{0}^{1} (1-x)^{m-1} dx = \sum_{m=1}^{n}  \frac{1}{m}  = H_n $$
A: We can also use binomial inversion.
We have (defining $H_0=0$) $$H_n = \sum\limits_{m=0}^{n} a_m\binom{n}{m},$$ where $a_0=0$ and $a_m = \frac{(-1)^{m-1}}{m}$ for all positive integers $m$. Hence by binomial inversion, we have
$$a_m = \sum\limits_{k=0}^{m} (-1)^{m+k}H_k\binom{m}{k}$$
for all non-negative integers $m$.
That is,
$$\frac{(-1)^{m-1}}{m} = \sum\limits_{k=0}^{m} (-1)^{m+k}H_k \binom{m}{k}$$
for all positive integers $m$ (by the definition of $a_m$).
This implies that
$$\frac{1}{m} = \sum\limits_{k=1}^{m} (-1)^{k-1}H_k \binom{m}{k}$$
(rememberring we defined $H_0=0$). Summing both sides over $m$ now gives the result.
