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Let $f_n(x):=n\left(e^{\frac{x^2}n}-1\right)$ on $\left[0,M\right]$. I believe this converges monotonically to $x^2$.

I used L'Hopital's rule to show pointwise convergence and the ratio of the derivatives I got converges monotonically to $x^2$ which I know implies convergence. But does it imply that the original $f_n$ also converge monotonically?

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Davide's answer is great. Another easy way is to substitute a Taylor series for $e^{x^2/n}$. – Jeff Nov 17 '11 at 21:00
@ Davide Giraudo I treated x as fixed and applied L'hopitals rule to the function of a real variable associated with the sequence [writing n as 1/(1/n)] – user9352 Nov 18 '11 at 12:49

We don't need L'Hospital's rule since we can write $$f_n(x)=n\int_0^{\frac{x^2}n}e^tdt=\int_0^{x^2}e^{\frac yn}dy,$$ making the substitution $y=nt$. Hence for a fixed $x$ the sequence $\{f_n(x)\}$ is decreasing. Furthermore, the convergence is uniform on all compact set $\left[-A,A\right]$ since $$\sup_{x\in \left[-A,A\right]}|f_n(x)-x^2|=\int_0^{x^2}\underbrace{(e^{\frac yn}-1)}_{\geq 0}dy\leq \int_0^{A^2}\left(\int_0^{\frac yn}e^tdt\right)dy,$$ and making the substitution $u=nt$, we get $$\sup_{x\in \left[-A,A\right]}|f_n(x)-x^2|\leq \int_0^{A^2}\left(\int_0^ye^{\frac un}\frac{du}n\right)dy\leq\frac 1n\int_0^{A^2}ye^{\frac yn}dy\leq \frac Ane^{\frac An}.$$ Since $f_n(\sqrt n)=n(e-1)$, we can't hope more.

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The power series for $e^t$ converges uniformly on closed (bounded) intervals, so the series for $e^{x^2/n}$ converges uniformly on $[0,M]$, uniformly in $n$. If you expand into a power series, subtract 1, and multiply by $n$, you end up with the uniformly convergent series $$ n(e^{x^2/n}-1) = x^2 + \sum_{k=2}^\infty \frac1{k!} \frac{x^{2k}}{n^{k-1}} $$ with all positive terms, each of which decreases in $n$ (except for $x^2$ itself). That proves the monotonicity you want - although Davide Giraudo's proof is better!

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