Prove that $$\sum_{r=1}^{n}(-1)^{r-1} \, H_{r} \, {}^{n} C_{r} = \frac{1}{n},$$ where ${}^{n}C_{k} = \binom{n}{k}$ and the Harmonic numbers are defined by $$H_{n} = \sum_{j=1}^{n} \frac{1}{j}.$$
My attempt: I opened the summations, and re-combined the series but I am getting the same terms again.
Using integration I was able to resolve the series and it converted into integration of $$ \int_{0}^{1} (1-x)^{n-1} \, dx.$$
$$\require{hyperref}
\begin{align}
\sum_{k=1}^n(-1)^{k-1}H_k\binom{n}{k}
&=\sum_{k=1}^n(-1)^{k-1}H_k\binom{n-1}{k-1}+\sum_{k=1}^n(-1)^{k-1}H_k\binom{n-1}{k}\tag1\\
&=\sum_{k=1}^n(-1)^{k-1}H_k\binom{n-1}{k-1}-\sum_{k=2}^{n+1}(-1)^{k-1}H_{k-1}\binom{n-1}{k-1}\tag2\\
&=\frac1n\sum_{k=1}^n(-1)^{k-1}\frac{n}{k}\binom{n-1}{k-1}\tag3\\
&=\frac1n\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\tag4\\[3pt]
&=\frac1n[n\gt0]\qquad\text{( where [...] are Iverson brackets)}\tag5
\end{align}
$$
Explanation:
$(1)$: $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$
$(2)$: substitute $k\mapsto k-1$ in the right sum
$(3)$: $H_k-H_{k-1}=\frac1k$
$(4)$: $\frac{n}{k}\binom{n-1}{k-1}=\binom{n}{k}$
$(5)$: if $n\gt0$, the sum including $k=0$ would be $-(1-1)^n=0$
$\phantom{\text{(5):}}$ so the sum without $k=0$ is $1$
$\phantom{\text{(5):}}$ if $n=0$, the original sum is empty
See Iverson brackets.
Using $$H_{n} = \int_{0}^{1} \frac{1- t^n}{1-t} \, dt$$ then \begin{align} \sum_{r=1}^{n}(-1)^{r-1} \, H_{r} \, \binom{n}{r} &= \int_{0}^{1} \left( \sum_{r=1}^{n} (-1)^{r-1} \binom{n}{r} \, (1 - t^r) \right)\, \frac{dt}{1-t} \\ &= \int_{0}^{1} \frac{1 - (1 - (1-t)^{n})}{1-t} \, dt \\ &= \int_{0}^{1} (1-t)^{n-1} \, dt\\ &= \frac{1}{n}. \end{align}
Notice that: $$\sum_{r=1}^{n} (-1)^{r-1} \binom{n}{r} \, t^r = 1 - \sum_{r=0}^{n} (-1)^{r} \binom{n}{r} \, t^r = 1 - (1 - t)^n$$ and when $t=0$ this yields, for $n \geq 1$, $$\sum_{r=1}^{n} (-1)^{r-1} \binom{n}{r} = 1 - \sum_{r=0}^{n} (-1)^{r} \binom{n}{r} = 1 - (1 - 1)^n = 1 - 0^n = 1.$$
