 Solution

The printed mean is $1/(\lambda i)$, so $T_i$ has exponential distribution of rate $i\lambda$. Let $S_k={\sum }_{i=1}^k{T}_i$, with fixed $k$. ThenHere is a direct proof, which also identifies the exceptional role of the slowest exponential clock. For $0<\theta <\lambda$, independence and the moment-generating function giveLetting $\theta \uparrow \lambda$ proves the tail upper bound $-\lambda x$. For a local lower bound at $x>0$, fix $0<\eta <x$ and $M>0$ with $\mathbb{P}({\sum }_{i=2}^k{T}_i\le M)>0$; for $k=1$ this probability is one. For large $N$, the eventputs $S_k/N$ in $(x-\eta ,x+\eta )$. Its logarithmic probability divided by $N$ tends to $-\lambda (x-\eta /2)$. Shrinking $\eta$ gives the local lower bound $-\lambda x$. At zero, $S_k/N\to 0$ almost surely and neighborhoods have probability tending to one.

For any closed set excluding zero, its nonnegative part is bounded away from zero; the tail upper bound applies at its infimum. Sets outside the support have probability zero. Together with the local lower bounds this proves the full large deviation principle, with a good rate function. Equivalently, the product large-deviation principle for the fixed vector $(T_1/N,\dots ,T_k/N)$ and the contraction principle for large deviations minimize ${\sum }_{i}i\lambda u_i$ subject to $u_i\ge 0$ and ${\sum }_i{u}_i=x$; all the excursion is assigned to $u_1$.