Probability of sum of exponentials r.v.'s greater than an exponential r.v. Let's say $T_1,T_2,...\sim\,$exp($\lambda$) and $S\sim\,$exp($\mu$), where $T_1,T_2,...$ are independent and independent of $S$. I want to calculate $P(\sum_{i=1}^nT_i>S)$. I don't want to solve the integral derectly, but to use the properties of independent exponential random variables. 
Here is my attempt:
\begin{align}
\dfrac{\mu}{n\lambda+\mu}=P(\min_{1\leq i\leq n}(T_i)>S)\leq P(\sum_{i=1}^nT_i>S)\leq \sum_{i=1}^nP(T_i>\dfrac{S}{n})=\dfrac{n\mu}{\frac{\lambda}{n}+\mu},
\end{align}
where I have used the properties (a) $P(T_i>S)=\dfrac{\mu}{\lambda+\mu}$ and (b) $\min_{1\leq i\leq n}(T_i)\sim\,$exp($n\lambda$).
¿Is there another way other than the integral using something like this?
 A: We can use a strong version of the memoryless property. Let $A,B$ be two independent, almost surely finite, positive random variables. For our purposes, we can assume they are absolutely continuous (they have densities $f_A$ and $f_B$). Assume they are also independent of $S$. Then,
\begin{align*}
P(S > A+B|S>A) &= \frac{P(S>A+B)}{P(S>A)} = \frac{\int_0^\infty P(S>B+t)f_A(t)dt}{\int_0^\infty P(S>t)f_A(t)dt}\\
&=\frac{\int_0^\infty P(S>B+t|S>t)P(S>t)f_A(t)dt}{\int_0^\infty P(S>t)f_A(t)dt}\\
&=\frac{\int_0^\infty P(S>B)P(S>t)f_A(t)dt}{\int_0^\infty P(S>t)f_A(t)dt}\\
&=\frac{P(S>B)\int_0^\infty P(S>t)f_A(t)dt}{\int_0^\infty P(S>t)f_A(t)dt}\\
&=P(S>B).
\end{align*}
Thus,
\begin{align*}
P\left(S>\sum_{i=1}^n T_i\right)&= P\left(S>\sum_{i=1}^n T_i\middle| S>\sum_{i=1}^{n-1} T_i\right)P\left(S>\sum_{i=1}^{n-1} T_i\right)\\
&=P\left(S>T_n\right)P\left(S>\sum_{i=1}^{n-1} T_i\right)\\
&= \frac{\lambda}{\lambda+\mu}P\left(S>\sum_{i=1}^{n-1} T_i\right).
\end{align*}
Unraveling the recursion, we get
$$P\left(S>\sum_{i=1}^n T_i\right) = \left(\frac{\lambda}{\lambda+\mu}\right)^n,$$
so
$$P\left(\sum_{i=1}^n T_i> S\right)= 1 - \left(\frac{\lambda}{\lambda+\mu}\right)^n.$$
