In a problem I'm working on, I'm asked to estimate, for each $n\in\mathbb{N}$, $\mathbb{P}(\sum_{k=1}^n X_n\geq 1)$ using Large Deviation where $X_1,X_2,\ldots$ are IID and uniformly distributed on $[0,1]$. From Large Deviation, we know that for each $n$ $$ \frac{1}{n}\log\left(\mathbb{P}\left(\sum_{k=1}^n X_n\geq na\right)\right)\leq -I(a) $$ where, denoting by log of the moment generating function $\psi(\lambda):=\log\mathbb{E}[e^{\lambda X_1}]$, $$ I(a):=\sup_{\lambda} \left(a\lambda - \psi(\lambda)\right). $$ We also know that, if $a$ is in the image of $\psi'$, then, by convexity, we can find a unique $\lambda_0(a)$ such that $I(a) = a\lambda_0(a) - \psi(\lambda_0(a))$.
To use these results in this problem, I set $a = \frac{1}{n}$ and have $$ \psi(\lambda) = \begin{cases} \log\left(\frac{e^{\lambda}-1}{\lambda}\right),&\quad \lambda\neq 0,\\ 0,&\quad \lambda= 0. \end{cases} $$
I was trying to compute $I(\frac{1}{n})$ and this is the part I'm stuck at. We can check that $\frac{1}{n}$ is in the image of $\psi'$ since the limits to $\pm\infty$ of $\psi'$ are $1$ and $0$; so $\lambda_0(\frac{1}{n})$ exists. However, I couldn't find a way to solve for such $\lambda_0(\frac{1}{n})$; I got $$ \frac{1}{n}=\psi'(\lambda_0(\textstyle\frac{1}{n}))= \frac{e^{\lambda_0(\frac{1}{n})}}{e^{\lambda_0(\frac{1}{n})}-1}-\frac{1}{\lambda_0(\frac{1}{n})}. $$ I was hoping that I'd get $I(\frac{1}{n})$ so that $e^{-nI(\frac{1}{n})}$ is roughly the same as how we compute the probability the usual way, which is $1-\frac{1}{n!}$.