Determine the domain of definition and the domain
Determine the domain of definition and the domain of continuity of $\displaystyle f(x)=\sum_{n=1}^{\infty}\frac1{n^2(1+n^2x^2)}$.
Explanation of Method
Dropping the non-negative term $n^2x^2$ from the denominator gives the immediate $x$-independent bound $1/n^2$, which is summable — an instant Weierstrass M-test that certifies uniform convergence, hence continuity, on all of $\mathbb R$ at once.
- Bound each term by $1/n^2$, uniformly in $x$.
- Apply the Weierstrass M-test to get uniform convergence on $\mathbb R$.
- Conclude continuity on $\mathbb R$ and note the function's asymptotic behavior.
Step 1 · Uniform bound
Since $1+n^2x^2\ge1$ for all $x$: $$ \frac1{n^2(1+n^2x^2)} \le \frac1{n^2} $$ for every $x\in\mathbb R$ and $n\ge1$.
Step 2 · Weierstrass M-test
Since $\sum_{n=1}^{\infty}1/n^2=\pi^2/6<\infty$, the M-test with $M_n=1/n^2$ gives uniform (and absolute) convergence on all of $\mathbb R$. In particular, the domain of definition is $(-\infty,\infty)$.
Step 3 · Continuity and asymptotics
Each term is continuous (denominator never $0$), so the uniformly convergent sum $f$ is continuous on all of $\mathbb R$. The function is even: $f(-x)=f(x)$. At $x=0$: $f(0)=\sum1/n^2=\pi^2/6$. For large $|x|$: $1/(n^2(1+n^2x^2))\approx1/(n^4x^2)$, so $f(x)\sim\zeta(4)/x^2=\pi^4/(90x^2)\to0$ as $|x|\to\infty$.
Summary Table
| Property | Value |
|---|---|
| Domain of definition | $(-\infty,\infty)$ |
| Domain of continuity | $(-\infty,\infty)$ |
| $f(0)$ | $\pi^2/6$ |
| Asymptotic as $|x|\to\infty$ | $f(x)\sim\pi^4/(90x^2)$ |
Additional Notes
- This series is the direct precursor of Problem 197's general result $\sum1/(n^2+[\varphi(x)]^2)$: writing $u=nx$, the term here is $\dfrac1{n^2}\cdot\dfrac1{1+(nx)^2}$, structurally the same drop-the-non-negative-term mechanism used throughout Problems 194, 197, and 198.
- Unlike Problems 206 and 207, where the domain of definition strictly contains or equals the continuity domain with a boundary issue, here both domains are simply all of $\mathbb R$ — no restriction is needed anywhere, since the majorant $1/n^2$ has no dependence on $x$ whatsoever.
- The function is differentiable as well (term-by-term differentiation is valid by the same M-test machinery applied to $f'(x)=-2x\sum1/(1+n^2x^2)^2$), making $f$ a smooth, bounded, even function that peaks at $x=0$ and decays like $1/x^2$.
Result
Defined and continuous on all of $\mathbb R$: