Expectation of a parallel system A system consists of $n$ components in parallel. The lifetimes of the components are i.i.d.
exp($\lambda$) random variables. The system functions as long as at least one of the $n$
components is functioning. Let $T$ be the lifetime of the system. Compute $E[T]$.

Ok, so I've done problems like these in the past and they usually involved a finite number of components with a fixed probability. That is, the components usually had a fixed probability $p$ of failure. So I would just plug that into a Binomial distribution and call it a day. This problem is a bit different because $p$ varies with time.
So I do have some ideas for this problem, but I eventually get stuck. 

Because this is a Poission process, the PP is counting the number of failures up to time $T$. So because the system fails when the last component fails, we essentially want to find the expected lifetime of the last component. In math terms:
Let $N(t)$ = the number of failures by time $t$ 
Let $T$ = the lifetime of the system
Let $S_i$ = the lifetime of component $i$
Let $X_i$ = the interarrival time between failure i and i-1$
So, essentially, we know that $T=max\{S_i | i \epsilon 1,2,...,n\}$
We also know that $E[T] = \sum_0^n E[S_i] $
So would the answe to this really be just the sum of all $\frac{1}{\lambda}$ or am I missing something?
 A: Due to the lack of memory property of the exponential distribution, when the $i$th failure happens, there are $n-i$ components left, whose residual lifetimes are still i.i.d. exponential of parameter $\lambda$. Hence the $(i+1)$th failure will occur at the time of the $i$th failure plus an exponential time of parameter $(n-i)\lambda$.
In other words, for every $1\leqslant i\leqslant n$, $X_i$ is exponential with parameter $(n-i+1)\lambda$ and $T=X_1+X_2+\cdots+X_n$ hence
$$
E(T)=\frac1\lambda\,\sum_{i=1}^n\frac1{n-i+1}=\frac1\lambda\,\sum_{k=1}^n\frac1k=\frac1\lambda\,H_n.
$$
Exercise: The naïve approach yields the alternative expression $E(T)=\frac1\lambda t_n$ where
$$
t_n=\sum_{k=1}^n{n\choose k}\frac{(-1)^{k+1}}k.
$$
One can show that $t_n=H_n$ thanks to a recursion on $n$ using the well known identities
$$
{n+1\choose k}={n\choose k}+{n\choose k-1},\quad{n\choose k-1}\frac1k=\frac1{n+1}{n+1\choose k},
$$
and
$$
\sum_{k=1}^{n+1}{n+1\choose k}(-1)^{k+1}=1.
$$
A: We first find the distribution of the maximum life $T$. The maximum of the $S_i$ is $\le t$ if and only if all the $S_i$ are $\le t$. This has probability $(1-e^{-\lambda t})^n$. Differentiate. The density function of $T$ is $\lambda n e^{-\lambda t}(1-e^{-\lambda t})^{n-1}$ (for $t\gt 0$). Call this $f_T(t)$ 
We want to find $\int_0^\infty tf_T(t)\,dt$. That is in principle no problem. Do the natural integration by parts, $u=t$, $dv=$ the rest. We end up needing to integrate $(1-e^{-\lambda t})^{n}$. Expand, using the Binomial Theorem. 
