Could someone help me accelerate the convergence of my series using Euler's transformation?

Question

I am trying to accelerate the convergence of the series. $$ \lim_{x \to \infty} \left(\sum_{n=0}^\infty(-1)^n \frac {x^{2n+1}}{(2n+1)n!}\right) $$ For the sake of simplicity, we can disregard the limit and look at the series as $$ \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)n!} $$ As the series is alternating I found that Euler's transformation could be of help, with the formula I found for it being: $$ \sum_{n=0}^\infty (-1)^n a_n =\sum_{n=0}^\infty (-1)^n \frac{(\Delta^n a)_0}{2^{2n+1}} $$ Where $$ (\Delta^n a)_0 = \sum_{k=0}^n (-1)^k \binom{n}{k} a_{n-k} $$ Apart from knowing that $$ a_n = \frac {x^{2n+1}}{(2n+1)n!} $$ I am honestly lost, and any help would be appreciated.

Answer 1

As I understand, it's needed to find a somewhat-quicker-converging approximation of $\sum_{n=0}^\infty(-1)^n \frac {x^{2n+1}}{(2n+1)n!}$ for large values of $x$. Despite it can be done by applying some transformation (as the one given here), from my point of view it will be rather useless.

Let's first consider the original power series, which is known to be the series expansion of $\varphi(x) = \int_0^{x} e^{-t^2/2} \mathrm{d} t$. Notice that $\lim_{x \to \infty} \left|\varphi(x)\right| = \left| \sqrt{\pi}/2\right|$, so $\varphi (x)$ is bounded (and also monotonically increasing). In what follows, let's consider only $x > 0$. From the properties of alternating series, for any given $x > 0$ and $N \in \mathbb{N}$ $$ \left|\varphi(x) - \sum_{n=0}^{N-1} (-1)^n a_n(x)\right| = \left|\varphi(x) - S_{N-1}(x)\right| \leq a_{N}(x), $$ where $a_n(x) = \frac {x^{2n+1}}{(2n+1)n!}$. For any given interval $[0, b]$ $$ \sup_{x\in[0, b]} \left|\varphi(x) - S_{N-1}(x)\right| \leq \frac{b^{2N+1}}{(2N+1)N!} \to 0, N \to \infty, $$ so for any given accuracy $\varepsilon > 0$ it is possible to find $N_{\varepsilon}$ such that $\sup_{x\in[0, b]} \left|\varphi(x) - S_{N-1}(x)\right| < \varepsilon$ for all $N \geq N_{\varepsilon}$, which means that approximation $S_{N-1}(x)$ will be "equally close" to $\varphi(x)$ for all $x \in [0, b]$ (which is, basically, the meaning of uniform convergence). However, one can check numerically, $N_{\varepsilon}$ grows as $b$ grows, so the larger is $[0, b]$, the more approximation terms are needed to ensure good uniform convergence.

But such approximation will be terrible for $x > b$, as $\lim_{x \to \infty} \left|S_N(x)\right| = +\infty$, so there exists $x_* > b$ such that $|S_N(x)| \geq \sqrt{\pi}$ for all $ x \geq x_*$ and $$ \left|\varphi(x) - C_N(x)\right| \geq \left|\left|C_N(x)\right| - \left|\varphi(x)\right|\right| \geq \left|C_N(x)\right| - \sqrt{\pi}/2. $$

However, if you need some computationally simple bounded approximation of $\varphi(x) = \int_0^x e^{-t^2/2} \mathrm{d} t$ for all $x \geq 0$, probability theory can be helpful. As $\Phi(x) = \frac{1}{2} + \frac{1}{\sqrt{2\pi}} \varphi(x)$ is the CDF of standard normal distribution, it can be approximated by CDF of finite sum of some random variables due to the Central Limit Theorem (CLT). A good candidate is CDF of the Irwin-Hall distribution, which is the distribution of sum of standard uniform random variables. It's already given in Wikipedia that $$ \Phi(x) \approx F_{IH}\left(\sqrt{\frac{12}{N}} x + \frac{N}{2}; N\right), $$ where $$F_{IH}(x; N) = \frac{1}{n!} \sum_{k=0}^{\lfloor x \rfloor} (-1)^k \binom{n}{k} (x-k)^n, \; 0 \leq x \leq N.$$ I'm not sure about specific bound on difference for this case (in general, rate of convergence in CLT is about $1 / \sqrt{N}$), but in practice $N = 12$ gives a very good approximation of $\Phi(x)$.