Putnam 2024, Problem B-6

For a real number \( a \), let \( F_a(x) = \sum_{n \geq 1} n^a e^{2n} x^{n^2} \) for \( 0 \leq x < 1 \). Find a real number \( c \) such that
\[ \lim_{x \to 1^-} F_a(x) e^{-1/(1-x)} = 0 \quad \text{for all } a < c, \text{ and} \]
\[ \lim_{x \to 1^-} F_a(x) e^{-1/(1-x)} = \infty \quad \text{for all } a > c. \]


Solution:
The claim holds with $c = -\frac{1}{2}$. Set $t := 1/(1-x)$, so that $x = 1 - 1/t$ and
$$-\frac{1}{t} - \frac{1}{t^2} \leq \log x \leq -\frac{1}{t}.$$

Suppose first that $a > -\frac{1}{2}$. Then
$$F_a(x)e^{-t} = \sum_{n=1}^{\infty} n^a e^{2n-t} x^{n^2}$$
$$\geq \sum_{n=1}^{\infty} n^a e^{2n-t-n^2/t-n^2/t^2}$$
$$= \sum_{n=1}^{\infty} n^a e^{-n^2/t^2} e^{-t(1-n/t)^2}.$$

If we restrict the sum to the range $t < n < t + \sqrt{t}$, we may bound the summand from below by $ct^a$; we then have $F_a(x)e^{-t} > ct^{a+1/2}$ and this tends to $\infty$ as $t \to \infty$.

Suppose next that $a < -\frac{1}{2}$. Then
$$F_a(x)e^{-t} = \sum_{n=1}^{\infty} n^a e^{2n-t} x^{n^2}$$
$$\leq \sum_{n=1}^{\infty} n^a e^{-t(1-n/t)^2}.$$

Fix $\varepsilon > 0$ such that $a + \varepsilon < -\frac{1}{2}$. For the summands with $t - t^{1/2+\varepsilon} < n < t + t^{1/2+\varepsilon}$, we may bound the summand from above by $ct^a$; this range of the sum is then dominated by $ct^{a+1/2+\varepsilon}$. For the summands with $n < t - t^{1/2+\varepsilon}$, we may bound the summand by $n^a e^{-t^{2\varepsilon}}$; this range of the sum is then dominated by $te^{-t^{2\varepsilon}}$. For the summands with $n > t - t^{1/2+\varepsilon}$, we may again bound the summand by $n^a e^{-t^{2\varepsilon}}$; this range of the sum is then dominated by $ct^{a+1}e^{-t^{2\varepsilon}}$. Since all three bounds tends to $0$ as $t \to \infty$, so then does $F_a(x)e^{-t}$.