Rapidly Decaying Exponential Series

April 10, 2026 · Luciano Muratore

1. Tail Estimate for ∑n=1∞e−n2\sum_{n=1}^{\infty}e^{-n^{2}}

The series ∑e−n2\sum e^{-n^{2}} converges extremely fast, meaning only a few terms are required to achieve high accuracy.

  • Tail beyond NN: The remainder RNR_N can be estimated as RN=∑n=N+1∞e−n2≈e−(N+1)2R_{N}=\sum_{n=N+1}^{\infty}e^{-n^{2}}\approx e^{-(N+1)^{2}}.
  • Example (N=5N=5): For N=5N=5, the tail is e−(5+1)2=e−36≈2.3×10−16e^{-(5+1)^{2}} = e^{-36} \approx 2.3 \times 10^{-16}.
  • Practicality: This value is essentially zero in double-precision arithmetic.

2. Uniform Convergence via Weierstrass M-test

Consider the parameterized series ∑n=1∞e−n2x\sum_{n=1}^{\infty}e^{-n^{2}x} where x>0x > 0[cite: 21, 22].

  • Comparison: Since n2≥nn^{2} \ge n for n≥1n \ge 1, it follows that e−n2x≤e−nxe^{-n^{2}x} \le e^{-nx}.
  • Geometric Series: The series ∑n=1∞e−nx=e−x1−e−x\sum_{n=1}^{\infty}e^{-nx} = \frac{e^{-x}}{1-e^{-x}} converges for x>0x > 0.
  • M-test Application: By setting Mn(x)=e−nxM_{n}(x) = e^{-nx}, we find that ∣e−n2x∣≤Mn(x)|e^{-n^{2}x}| \le M_{n}(x).
  • Conclusion: Because ∑Mn(x)\sum M_{n}(x) converges, the series ∑e−n2x\sum e^{-n^{2}x} converges uniformly on the interval [a,∞)[a, \infty) for any a>0a > 0.

3. The Sequence e−n2e^{-n^{2}} in lpl^p Spaces

For any p>0p > 0, the lpl^p norm of the sequence (e−n2)(e^{-n^{2}}) is defined by: ∥e−n2∥pp=∑n=1∞(e−n2)p=∑n=1∞e−pn2\|e^{-n^{2}}\|_{p}^{p} = \sum_{n=1}^{\infty}(e^{-n^{2}})^{p} = \sum_{n=1}^{\infty}e^{-pn^{2}}.

  • Decay Rate: The terms e−pn2e^{-pn^{2}} decay super-exponentially.
  • Convergence: Since ∑n=1∞e−pn2<∞\sum_{n=1}^{\infty}e^{-pn^{2}} < \infty for all p>0p > 0, the sequence is a member of every lpl^p space.
  • Result: (e−n2)n=1∞∈⋂p>0lp(e^{-n^{2}})_{n=1}^{\infty} \in \bigcap_{p>0} l^{p}.

4. Dominance and lpl^p Membership

A sequence (bn)(b_n) can be guaranteed lpl^p membership if it is dominated by the rapidly decaying exponential e−n2e^{-n^2}.

Key Observation

If there exists a constant C>0C > 0 such that ∣bn∣≤Ce−n2|b_{n}| \le Ce^{-n^{2}}: ∑n=1∞∣bn∣p≤Cp∑n=1∞e−pn2<∞\sum_{n=1}^{\infty}|b_{n}|^{p} \le C^{p}\sum_{n=1}^{\infty}e^{-pn^{2}} < \infty.

Important Distinction

  • Multiplicative Dominance: ∣bn∣≤Ce−n2⇒bn∈⋂p>0lp|b_{n}| \le Ce^{-n^{2}} \Rightarrow b_{n} \in \bigcap_{p>0} l^{p}.
  • Additive Inclusion (Counterexample): If a sequence is formed by bn=e−n2+1nb_{n} = e^{-n^{2}} + \frac{1}{n}, it does not belong to l1l^1 because the 1n\frac{1}{n} term dominates the tail behavior.