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The function $x{e}^{-x}$ is strictly increasing on $(0,1)$ and strictly decreasing on $(1,\infty )$. Hence for each $\lambda >1$ there is a unique $\theta ={\lambda }^{\ast }\in (0,1)$ with $\theta e^{-\theta }=\lambda e^{-\lambda }$.

Apply the subcritical component estimates to $G(n,\theta /n)$. For a specified vertex, the exact tree-component expectation in the Erdős-Rényi model gives the limiting probability that it lies in a tree component of order $j$:For each fixed $j$, the probability of a non-tree graph component of that size is $O_j(1/n)$: a spanning tree and one extra edge are necessary. Moreover the Galton-Watson process exploration estimate from the preceding proof bounds the probability of order greater than $K$ by $e^{-{\delta }_{\theta }(K-1)}$, uniformly in $n$, where ${\delta }_{\theta }=\theta -1-\log \theta >0$. First let $n\to \infty$ for fixed $K$, then let $K\to \infty$. No probability mass escapes to large components or non-tree components, so ${\sum }_{j\ge 1}{a}_j=1$. Multiplying by $\theta$ and using its defining identity givesEquivalently, the rooted-tree generating function selects the smaller solution of $T{e}^{-T}=\lambda e^{-\lambda }$. The choice of that branch is essential; the larger solution $\lambda$ is not the value of this convergent series.

Assume $0<\lambda =1-\epsilon <1$, as required for this nontrivial subcritical binomial random graph. Put $\delta =\lambda -1-\log \lambda >0$. A breadth-first exploration of a binomial random graph is dominated by a Galton-Watson process whose offspring counts are independent random variables with law $\mathrm{Bin}(n,\lambda /n)$. If its total progeny $T$ is at least $j$, the first $j-1$ offspring counts sum to at least $j-1$. Their moment-generating function is bounded by that of a Poisson distribution of mean $\lambda (j-1)$. The exponential Markov inequality, optimized at $t=-\log \lambda$, yieldsTaking $j=\lceil (1+\eta )\log n/\delta \rceil$ for any fixed $\eta >0$, a union bound over all starting vertices gives $\mathbb{P}(L_1\ge j)=o(1)$.

For the matching lower bound, let $X_j$ count tree components of order $j$. The tree-component expectation in the Erdős-Rényi model is exactlyFor $j=O(\log n)$, the Stirling formula givesChoose $j=\lfloor (1-\eta )\log n/\delta \rfloor$, with $0<\eta <1$. Then $\mathbb{E}{X}_j\to \infty$. Distinct overlapping vertex sets cannot both be graph components. For disjoint sets, the joint probability differs from the product only because the $j^2$ between-set edges were counted twice as absent. ThusIt follows that $\mathrm{var}{X}_j/(\mathbb{E}{X}_j{)}^2=o(1)$. The second moment method gives $X_j>0$ with high probability. Combining both bounds, and then taking arbitrarily small fixed $\eta$, proves the sharp subcritical largest-component scaleFinally the Taylor expansion of $-\log (1-\epsilon )$ gives $\delta ={\epsilon }^2/2+{\epsilon }^3/3+\cdots$. The endpoint $\lambda =0$ has no edges and is excluded from this logarithmic asymptotic.

The exponential generating function for labelled rooted trees satisfies $T=z{e}^T$: remove the root and obtain an unordered set of smaller labelled rooted trees. For $0\le z<1/e$, the convergent series is the solution in $[0,1)$ of $T{e}^{-T}=z$, rather than the larger real solution. Alternatively, tree-component expectation in the Erdős-Rényi model and subcritical exploration give $T(\theta e^{-\theta })=\theta$ for $0<\theta <1$: their limiting probabilities for the order of the tree component of a uniform vertex sum to one. The component tail bound makes this passage through the infinite sum valid.

For fixed $0<\lambda <1$ in $G(n,\lambda /n)$, the largest component of a graph has the displayed order with high probability. A dominating Galton-Watson process gives the tail bound $\mathbb{P}(T\ge j)\le e^{-(\lambda -1-\log \lambda )(j-1)}$ by the exponential Markov inequality. For the lower bound, the tree-component expectation in the Erdős-Rényi model diverges at $j=\lfloor (1-\eta )\log n/(\lambda -1-\log \lambda )\rfloor$. The disjoint-set covariance factor and the exclusion of overlaps give relative variance tending to zero, so the second moment method supplies a tree component of that order.