I'm studying for an upcoming exam in analysis and am focusing on the following problem:
Prove that the limit $$ \lim_{n \rightarrow \infty} \int_0^1 \bigg[ \sum_{k=0}^{n} \frac{x^k}{k!} \bigg]^2 dx$$ exists.
I'm looking for some guidance on my proof - I feel justified in what I've come up with, but I can't seem to verify the result in Matlab, so I think I must have gone wrong somewhere. My attempt:
Consider $f_n:[0,1] \rightarrow \mathbb{R}$, defined as $f_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!}$. We claim that $f_n(x)$ converges uniformly to $f(x) = e^x$ on $[0,1]$. Accordingly, since $$ \sum_{k=0}^{\infty} \frac{x^k}{k!} = e^x \qquad \forall x \in \mathbb{R}$$ we have that for any $\epsilon >0$, $\exists N$ such that $$ |f_n(x) - f(x)| < \epsilon $$ provided that $n \geq N$. Since this holds for all $x \in \mathbb{R}$, it certainly holds for $x \in [0,1]$, thus we have the desired uniform convergence. Now, since $f_n(x)$ is a sequence of continuous functions ($f_n(x)$ is a polynomial for each $n$), it must hold that $g_n(x) = [f_n(x)]^2$ converges uniformly to $g(x) = f(x)^2 = e^{2x}$ on $[0,1]$ (proved in an earlier problem). Finally, since the convergence of the integrand $g_n(x)$ is uniform, and each $g_n(x)$ is continuous, thus integrable, we have that \begin{align*} \lim_{n \rightarrow \infty} \int_0^1 \bigg[ \sum_{k=0}^{n} \frac{x^k}{k!} \bigg]^2 dx &= \int_0^1 e^{2x} dx \\ &= \frac{1}{2} (e^2 - 1). \\ \end{align*} Our result is finite, thus the limit exists.
Have I made a mistake at some point?
As noted, the proof is wrong since the convergence of the series definition of the exponential to $e^x$ is not uniform on $\mathbb{R}$. This may help: $$ \left|\left(\sum_{k=0}^n \frac{x^k}{k!}\right)^2-e^{2x} \right|\\ =\left|\sum_{k=0}^n \frac{x^k}{k!}-e^{x} \right|\left|\sum_{k=0}^n \frac{x^k}{k!}+e^{x} \right|\\ $$ the first term in the product can be kept uniformly small on $[0,1]$ (see Weirstrass-M test, on this interval, the convergence is uniform). The second is bounded by $e^2$, since $$ \forall n\in \mathbb{N},\;\sum_{k=0}^n \frac{x^k}{k!}\leq e^x $$ so $$ \sup_{x\in[0,1]}\left\{ \sum_{k=0}^n \frac{x^k}{k!}+e^{x}\right\} \leq e^2 $$ and you have uniform convergence and may interchange limit and integral and perform a simple integration to find the limit is in fact $$ \frac12 (e^2-1) $$