Prove that these two series are equal

Question

I am doing a course in Reliability. There is a system of n parallel identical components, each of which has a failure probability $$exp(-\lambda t)$$. As is the case with parallel systems, the system works if at least one component works OR, the system fails if all the components fail. Therefore, R(t) be the probability that the system works. $$\therefore R(t) = 1 - (1 - e^{-\lambda t} )^n - (1) $$. i.e, probability that no components fail. Also, $$R(t)= \sum_{x=1}^n {^nC_x}(e^{-\lambda t})^x(1-e^{-\lambda t})^{n-x} - (2)$$ Now, it is easy to see that (1) and (2) are same. Now, we are to calculate the integral $$\int_0^\infty R(t)dt$$ (which is actually called the mean life time of the system). Now from (1) $$\int_0^\infty R(t)dt = \int_0^\infty (1 - (1 - e^{-\lambda t} )^n )dt = \int_0^\infty \sum_{x=1}^n {^nC_x}\times(-1)^{i+1}\times e^{-x \lambda t} (Binomial Expansion) = \frac{1}{\lambda} \sum_{x=1}^n {^nC_x \frac{(-1)^{x+1}}{x}} -(a) $$ Also, from (2) $$ \int_0^\infty R(t)dt= \int_0^\infty \sum_{x=1}^n {^nC_x}(e^{-\lambda t})^x(1-e^{-\lambda t})^{n-x} dt = \frac{1}{\lambda} \int_0^1 \sum_{x=1}^n {^nC_x}v^{x-1}(1-v)^{n-x}dv$$ (taking transformation) $$v = e^{-\lambda t}$$. Thus it becomes, $$\frac{1}{\lambda} \int_0^1 \sum_{x=1}^n{(^nC_x)}v^{x-1}(1-v)^{\bigl(n-(x-1)\bigr)-1}dv = \frac{1}{\lambda} \sum_{x=1}^n {(^nC_x)}\frac{\Gamma(x)\Gamma(n-x+1)}{\Gamma(n+1)} = \frac{1}{\lambda} \sum_{x=1}^n \frac{1}{x} - (b) $$ Now, (a) and (b) are derived from same expression, so they are same. You can also check by taking any arbitrary integer value of n. But what is bothering me it is not directly obvious that expressions (a) and (b) are same. Can anyone give any direct proof that series (a) and series (b) are same, i.e $$\sum_{x=1}^n \frac{1}{x} = \sum_{x=1}^n {^nC_x \frac{(-1)^{x+1}}{x}} $$ ?

Answer 1

Alternatively, consider the generating function of the RHS: \begin{align} \sum_{n=1}^\infty \sum_{k=1}^n \frac{(-1)^{k+1}}{k}\binom{n}{k} z^n &=-\sum_{k=1}^\infty \frac{(-z)^k}{k} \sum_{n=k}^\infty \binom{n}{k} z^{n-k} \\ &=-\sum_{k=1}^\infty \frac{(-z)^k}{k} \cdot\frac{1}{(1-z)^{k+1}} \\ &=-\frac{1}{1-z}\sum_{k=1}^\infty \frac{1}{k}\left(\frac{-z}{1-z}\right)^k \\ &=\frac{1}{1-z} \log\left(1-\frac{-z}{1-z}\right) \\ &=\frac{1}{1-z} \log\left(\frac{1}{1-z}\right) \\ &=-\frac{\log(1-z)}{1-z}, \end{align} which is known to be the generating function of the harmonic numbers $H_n$, so $$\sum_{k=1}^n \frac{(-1)^{k+1}}{k}\binom{n}{k}=H_n.$$