Find the domain of definition of f(x)=sum from n=1

Find the domain of definition of $\displaystyle f(x)=\sum_{n=1}^{\infty}\left(\frac{n+1}n\right)^{-n^2}e^{n\frac{1+x}{1-x}}$.

Explanation of Method

Applying the root test directly to the general term combines the coefficient's $e^{-n}$-type decay (from $(1+1/n)^{-n^2}$) with the exponential factor, collapsing to a single condition $e^{2x/(1-x)}<1$. Solving this inequality algebraically, then checking the resulting boundary point $x=0$ separately, gives the full domain.

Apply the root test to the full term $a_n(x)$.
Solve the resulting inequality $\frac{2x}{1-x}<0$.
Check the boundary point $x=0$ directly, since the root test is inconclusive there.

Step 1 · Root test on the full term

With $a_n(x)=\left(\dfrac n{n+1}\right)^{n^2}e^{n\frac{1+x}{1-x}}$: $$ \sqrt[n]{a_n(x)} = \left(\frac n{n+1}\right)^n \cdot e^{\frac{1+x}{1-x}} \longrightarrow \frac1e\cdot e^{\frac{1+x}{1-x}} = e^{\frac{1+x}{1-x}-1} = e^{\frac{2x}{1-x}}, $$ using $\left(\frac n{n+1}\right)^n=\left(1+\frac1n\right)^{-n}\to e^{-1}$ and simplifying $\frac{1+x}{1-x}-1=\frac{2x}{1-x}$. The series converges when this limit is $<1$: $$ e^{\frac{2x}{1-x}}<1 \iff \frac{2x}{1-x}<0. $$

Step 2 · Solve the inequality

$\dfrac{2x}{1-x}<0$ iff $x$ and $1-x$ have opposite signs: $$ (x>0 \text{ and } 1-x<0) \implies x>1, \qquad (x<0 \text{ and } 1-x>0) \implies x<0. $$ So the root test gives convergence for $x<0$ or $x>1$, and divergence for $0

Step 3 · The boundary point $x=0$

At $x=0$: $\frac{2x}{1-x}=0$, so the root test limit is exactly $e^0=1$ — inconclusive. Checking the term directly: at $x=0$, $\frac{1+x}{1-x}=1$, so $$ a_n(0) = \left(\frac n{n+1}\right)^{n^2}e^{n}. $$ Using the refined asymptotic $\left(\frac n{n+1}\right)^{n^2}\sim\dfrac{e^{-n}}{\sqrt e}$ (Problem 233): $$ a_n(0) \sim \frac{e^{-n}}{\sqrt e}\cdot e^n = \frac1{\sqrt e} \ne 0. $$ Since the terms do not tend to $0$, the series diverges at $x=0$ by the Divergence Test.

Summary Table

Region	Behavior
$x<0$	Converges
$x=0$	Diverges (terms $\to1/\sqrt e\ne0$)
$0	Diverges (root test, or undefined at $x=1$)
$x>1$	Converges

Additional Notes

This problem directly reuses the refined asymptotic $(1+1/n)^{n^2}\sim e^n/\sqrt e$ derived in Problem 233 — the boundary point $x=0$ here is exactly analogous to that problem's boundary $x=1/e$, both requiring the second-order correction term to resolve a root-test tie.
The domain splits into two disjoint unbounded pieces, $(-\infty,0)$ and $(1,\infty)$, separated by the "forbidden zone" $[0,1]$ where the series either diverges outright or (at $x=1$) is not even defined — a genuinely different topology from the single-interval domains seen in Problems 223–238.
The substitution $\frac{1+x}{1-x}$ is a Möbius transformation mapping $x=0\mapsto1$, $x=1\mapsto\infty$, and $x=-1\mapsto0$; the exponent $e^{n\cdot(\cdot)}$ then translates the transformed variable's sign directly into convergence or divergence via the root test.

Result

$$ \text{Domain of } f(x): \quad x\in(-\infty,0)\cup(1,\infty). $$