Test the uniform convergence of sum from n=1 to infinity
Test the uniform convergence of $\displaystyle\sum_{n=1}^{\infty}x^{2n}e^{-n^2x^2}$ on $x\in(-\infty,\infty)$.
Explanation of Method
Unlike a plain power series, this term's coefficient itself decays in $x$ (via the Gaussian factor), so each term achieves a genuine finite maximum over all of $\mathbb R$. Finding that maximum by calculus gives an $x$-independent bound $M_n$ that decays faster than any geometric series, so the Weierstrass M-test applies on the whole real line.
- Maximize $u_n(x)=x^{2n}e^{-n^2x^2}$ over $x\in\mathbb R$ using calculus.
- Bound the resulting maximum $M_n=(ne)^{-n}$ by a convergent geometric majorant.
- Apply the Weierstrass M-test to conclude uniform convergence on $\mathbb R$.
Step 1 · Maximize $u_n(x)$
Let $t=x^2\ge0$, so $u_n=t^ne^{-n^2t}$. Then $\dfrac{d}{dt}\left[t^ne^{-n^2t}\right]=nt^{n-1}e^{-n^2t}(1-nt)$, which vanishes (for $t>0$) at $t=1/n$, i.e. $x=\pm1/\sqrt n$. Since $u_n=0$ at $t=0$ and $u_n\to0$ as $t\to\infty$, this interior critical point is the maximum.
Step 2 · Evaluate the maximum
At $t=1/n$: $$ u_n\!\left(\frac1{\sqrt n}\right) = \left(\frac1n\right)^n e^{-n^2/n} = \frac1{n^n}\cdot e^{-n} = \frac1{(ne)^n}. $$ So $\displaystyle\sup_{x\in\mathbb R}|u_n(x)|=M_n=\frac1{(ne)^n}$.
Step 3 · Majorant series and the M-test
Since $n\ge1$, $ne\ge e>1$, so $(ne)^n\ge e^n$, giving $M_n=\dfrac1{(ne)^n}\le\dfrac1{e^n}$. Since $\sum_{n=1}^{\infty}e^{-n}$ is a convergent geometric series (ratio $1/e<1$), $\sum M_n<\infty$. By the Weierstrass M-test with this majorant, the series converges uniformly and absolutely on all of $\mathbb R$.
Summary Table
| Quantity | Value |
|---|---|
| Maximizer | $x=\pm1/\sqrt n$ |
| $M_n=\sup_x u_n(x)$ | $1/(ne)^n\le e^{-n}$ |
Additional Notes
- The technique of finding the exact supremum of each term via calculus (rather than a crude algebraic bound) is what makes the M-test bite here; a naive bound like $|x^{2n}|\le1$ would fail since $x$ ranges over all of $\mathbb R$, but the Gaussian factor caps each term's true maximum at a value that shrinks super-exponentially in $n$.
- This "moving maximizer" phenomenon — the peak location $x=1/\sqrt n\to0$ shifts toward the origin as $n$ grows — is structurally similar to Problem 195's moving-bump counterexample, except here the peak height $(ne)^{-n}$ decays fast enough to still guarantee uniform convergence, unlike Problem 195 where the peak heights stayed bounded away from $0$.
- Since the function is even ($u_n(-x)=u_n(x)$) and each term is continuous, the uniformly convergent sum $f(x)=\sum x^{2n}e^{-n^2x^2}$ is an even, continuous function on $\mathbb R$ with $f(0)=0$ and $f(x)\to0$ as $|x|\to\infty$.
Result
Converges uniformly and absolutely on all of $\mathbb R$: