A "k out of n" system functions if, as the name suggests, k of the n components of the system remain operational, i.e. the system fails upon failure of any n-k+1 components.
My A/S/M MAS-I study manual (1st edition, 4th print) states without proof: for a k out of n system with components having identical exponentially distributed survival times with mean θ, expected time to failure is
$$ \theta \sum_{i=k}^n \frac{1}{i} $$
I'm stuck on proving this. A/S/M doesn't really cite sources and my searches on this site and others have come up short. I appreicate the help!
One of the reading materials for the exam provided me the answer: Introduction to Probability Models by Sheldon M. Ross, p.590 http://files.kevinl.io/introduction-to-prob-models-11th-edition.PDF
The exponential distribution gives distribution function $F(t)= 1 - e^{-t/\theta}$ and survival function $S(t)=e^{-t/\theta}$ for each component, so we have a joint survival function for the system given by
$$S(t)={\sum_{i=k}^n}\binom{n}{i}(e^{-t/\theta})^{i}(1-e^{-t/\theta})^{n-i}$$
The support of our distribution is non-negative so we can use ${E(t)=\int_{0}^\infty} S(t)dt$ to calculate the expected time to failure. Thus
$$E(t)={\int_{0}^\infty}{\sum_{i=k}^n}\binom{n}{i}(e^{-t/\theta})^{i}(1-e^{-t/\theta})^{n-i}dt$$ $$={\sum_{i=k}^n}\binom{n}{i}{\int_{0}^\infty}(e^{-t/\theta})^{i}(1-e^{-t/\theta})^{n-i}dt$$
Substituting $u=e^{-t/\theta}$, $du=-\frac{1}{\theta}e^{-t/\theta}dt$ gives
$$E(t)=\theta{\sum_{i=k}^n} \binom{n}{i}{\int_{0}^1}u^{i-1}(1-u)^{n-i}dt$$
Embarrassingly this is where I got stuck but letting $C(y;a,b)=\int_{0}^1y^a(1-y)^bdy$, it is easily shown that $C(y;a,0)=\frac{1}{a+1}$, and via integration by parts $C(y;a,b)=\frac{b}{a+1}C(y;a+1,b-1)$. From these two relations induction gives $C(y;a,b)=\frac{a!b!}{(a+b+1)!}$.
Alternatively it could be noted that $f(y)=\frac{\Gamma(a+b+2)}{\Gamma(a+1)\Gamma(b+1)}y^a(1-y)^b$ is the density function of a Beta distribution with positive integer parameters a+1 and b+1, and since we must have $\int_{0}^1f(y)dy=1$, then $\int_{0}^1C(y;a,b)dy=\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}=\frac{a!b!}{(a+b+1)!}$.
In either case we can now produce
$$E(t)=\theta{\sum_{i=k}^n} \binom{n}{i}{\int_{0}^1}u^{i-1}(1-u)^{n-i}dt$$ $$=\theta{\sum_{i=k}^n}\frac{n!}{i!(n-i)!}\frac{(i-1)!(n-i)!}{n!}$$ $$=\theta{\sum_{i=k}^n}\frac{1}{i}$$