From the memoryless property of Exponential Distribution function:
Suppose E(1,λ) and E(1,μ) are independent, then;
P [min{ E(1,λ) , E(1,μ) } > t] = P [E(1,λ) > t] . P [E(1,μ) } > t]
= exp (-λt) . exp (-μt)
= exp {-(λ + μ)t}
which shows that minimum of E(1,λ) and E(1,μ) is exponentially distributed.
So,
E(1, λ1+ λ2+ λ3+……. λn) = min { E(1,λ1), E(1,λ2), E(1,λ3), ……….., E(1,λn) }
Here, if we put λ = 1, then;
E(1, 1+ 2+ 3+……. n) = min { E(1,1), E(1,2), E(1,3), ……….., E(1,n) }
