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I have been given the following series: $$ \sum_{n=1}^\infty e^{-n^2x^2} $$ Let now $a>0$. Argue that the series converge uniformly on the interval $[a,\infty)=\{x\in\mathbb{R}:x\geq a\}$. To do this i have been using Weierstrass' M-test. First i have said that as the exponential function grows faster than the power function the following should be true: $ \frac{1}{e^{n^2x^2}}\leq \frac{1}{n^2} $. We know that $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent which makes it a convergent majorant series. Pr. Weierstrass' M-test the series $\sum_{n=1}^\infty e^{-n^2x^2}$ is then uniformly convergent. Is this approach okay?

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You have to prove that your upper bound is valid. Use the fact that $e^{x} \geq x$ for $x >0$ (which follows from the Taylor expansion). Now $e^{n^{2}x^{2}} \geq e^{n^{2}a^{2}} \geq n^{2}a^{2}$ so $\frac 1 {e^{n^{2}x^{2}}} \leq \frac 1 {n^{2}a^{2}}$, now compare with the series $ \sum \frac 1 {n^{2}a^{2}}$ which is convergent.

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