Cumulative distribution function of normal distribution as a series According to Wikipedia the cumulative distribution function of the standard normal distribution can be approximated with the following series:
$\frac12 + \frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}} [x + \frac{x^3}{3} + \frac{x^5}{{3} * {5}} + ... + \frac{x^{2n+1}}{(2n+1)!!} + ...]$
Can a series be used to approximate a normal distribution if $mu$ is not 0 and/or $\sigma$ is not 1?
Here is an R function that implements the above series for the standard normal.  However, I cannot figure out how to modify this series if $mu$ is not 0 and/or $\sigma$ is not 1:
my.cdf.function <- function(x) {
     n <- 201
     my.seq <- seq(1,n,by=2)
     my.cdf.series <- sum(sapply(1:length(my.seq), function(i) {x^my.seq[i] / cumprod(my.seq[1:i])[i]}))
     cdf.x = 1/2 + 1/((2 * pi)^0.5) * (exp(1)^((-x^2)/2)) * my.cdf.series
     return(cdf.x)
}

my.data <- data.frame(x = seq(-5,5,by=0.01))
my.output <- apply(my.data, 1, my.cdf.function)

pnorm.cdf <- pnorm(my.data$x, mean = 0, sd = 1)

all.equal(round(my.output,8), round(pnorm.cdf,8))
#[1] TRUE

plot(my.data$x, my.output)


EDIT
Here is an alternative approach, also from Wikipedia, that works for the standard normal, but I have not been able to get it to work if $mu$ is not 0 and/or $\sigma$ is not 1:
$\Phi(x) = \frac12 + \frac{1}{\sqrt{2\pi}} \sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{2^k k! (2k+1)}$
Here is R code:
x1 <- -5.5
x2 <-  4
sigma <- 1
mu    <- 0
n <- 0:200
y1 <- 0.5+(1/((2*pi)^0.5))*sum((((-1)^n)*(x1^(2*n+1)))/((2^n)*(2*n+1)*factorial(n)))
y2 <- 0.5+(1/((2*pi)^0.5))*sum((((-1)^n)*(x2^(2*n+1)))/((2^n)*(2*n+1)*factorial(n)))
y2-y1
#[1] 0.9999683

pnorm(4, 0, 1)
#[1] 0.9999683

 A: @John L essentially answered my question in a comment.  He wrote: "To find $P[X<x]$ for any normally distributed random variable with mean $μ$ and standard deviation $σ$, start by calculating $y=\frac{x−μ}{σ}$, then evaluate the standard normal distribution function, $Φ(y)$. You are done."
I had been confused erroneously thinking I also needed to use $\frac{1}{\sigma\sqrt{2\pi}}$.
Here is modified R code that implements John's approach.
# cdf of normal

mu <- 2
sigma <- 2

my.cdf.function <- function(x) {
     y <- (x - mu) / sigma
     n <- 201
     my.seq <- seq(1,n,by=2)
     my.cdf.series <- sum(sapply(1:length(my.seq), function(i) {y^my.seq[i] / cumprod(my.seq[1:i])[i]}))
     cdf.x = 1/2 + 1/((2 * pi)^0.5) * (exp(1)^((-y^2)/2)) * my.cdf.series
     return(cdf.x)
}

my.data <- data.frame(x = seq(-5.5,10,by=0.01))
my.output <- apply(my.data, 1, my.cdf.function)

pnorm.cdf <- pnorm(my.data$x, mean = 2, sd = 2)

all.equal(round(my.output,8), round(pnorm.cdf,8))
#[1] TRUE

jpeg(file="myplot3.jpeg")
plot(my.data$x, my.output)
dev.off()


