A question on binomial theorem If $C_0$, $C_1$, $C_2$,...$C_n$ are the coefficients in the expansion of $(1+x)^n$, where $n$ is a positive integer, show that $$C_1- {C_2\over 2}
+{C_3\over 3}-...+{(-1)^{n-1} C_n\over n}=1+ {1\over 2}+ {1\over 3}+...+{1\over n}$$

In other words, you are saying that
\begin{equation}
\sum_{k=1}^n \dfrac{\left(-1\right)^{k-1}}{k} \dbinom{n}{k} = \dfrac{1}{1} + \dfrac{1}{2} + \cdots + \dfrac{1}{n}
\end{equation}
(because your $C_k$ are precisely the binomial coefficients $\dbinom{n}{k}$).
This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.
(If you're looking for a hint: Induction on $n$.)
