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Write $S_N={\sum }_{i=1}^N{T}_i$. The answer isThe sum of exponential variables with linearly increasing rates has the distribution of the maximum of $N$ independent rate-$\lambda$ exponential random variables. To prove this identity, start $N$ independent rate-$\lambda$ exponential clocks. The first rings after a rate-$N\lambda$ exponential time; the memoryless property leaves $N-1$ fresh independent clocks. Repeating this argument shows that the successive order statistic gaps are independent, with rates $N\lambda ,(N-1)\lambda ,\dots ,\lambda$. Their sum has the distribution of $S_N$. ThereforeFor $x>0$, $N{e}^{-\lambda Nx}\to 0$, and henceIf $0<u<v$, subtracting the two tails gives the interval exponent $-\lambda u$, since the tail at $Nv$ is exponentially smaller. Also $\mathbb{E}{S}_N={\lambda }^{-1}{\sum }_{i=1}^N{i}^{-1}=O(\log N)$, so the Markov inequality gives $S_N/N\to 0$ in probability. Local lower bounds at positive points and at zero follow; the tail estimate gives the closed-set upper bound exactly as in part (a). This proves the full large deviation principle.

There is also a deduction from part (b). For fixed $k$ and $i>k$, a rate-$i\lambda$ exponential random variable is stochastically smaller than a rate-$k\lambda$ one. Coupling by the same uniform variables therefore gives $S_N\le S_k+Z_N$ with independent additional copies. For fixed $x>0$ and sufficiently large $k$, the upper-tail exponent from part (b) is $-\lambda x-\log ((k-1)/k)$. Let $k\to \infty$ to obtain $-\lambda x$; the lower tail estimate $\mathbb{P}(S_N>Nx)\ge \mathbb{P}(T_1>Nx)=e^{-\lambda Nx}$ matches it. The exact maximum identity additionally supplies all local lower bounds.