How would I show that the error function erf is strictly increasing?

I got the Taylor series:
$$\sum _{ n=0 }^{ \infty }{ \frac { (-1)^{ n }x^{ 2n+1 } }{ n!(2n+1) } } $$

and I applied $$\frac{a_{n+1}}{a_n} 1$$ however it does not work out to be greater than 1.

$$\frac { a_{ n+1 } }{ a_{ n } } =\frac { (-1)^{ n+1 }x^{ 2(n+1)+1 } }{ (n+1)!(2(n+1)+1) } \cdot \frac { n!(2n+1) }{ (-1)^{ n }x^{ 2n+1 } } =\frac { (-1)(-1)^{ n }x^{ 2n }x^{ 3 } }{ n!(n+1)(2n+3) } \cdot \frac { n!(2n+1) }{ (-1)^{ n }x^{ 2n } x } $$
$$==\frac { (-1)(2n+1) }{ (n+1)(2n+3) }x^{ 2 }\ngtr 1$$

Question

How would I show that the error function erf is strictly increasing? 

I got the Taylor series:
$$\sum _{ n=0 }^{ \infty  }{ \frac { (-1)^{ n }x^{ 2n+1 } }{ n!(2n+1) }  } $$

and I applied $$\frac{a_{n+1}}{a_n}  > 1$$ however it does not work out to be greater than 1. 

$$\frac { a_{ n+1 } }{ a_{ n } } =\frac { (-1)^{ n+1 }x^{ 2(n+1)+1 } }{ (n+1)!(2(n+1)+1) } \cdot \frac { n!(2n+1) }{ (-1)^{ n }x^{ 2n+1 } } =\frac { (-1)(-1)^{ n }x^{ 2n }x^{ 3 } }{ n!(n+1)(2n+3) } \cdot \frac { n!(2n+1) }{ (-1)^{ n }x^{ 2n } x } $$
$$==\frac { (-1)(2n+1) }{ (n+1)(2n+3) }x^{ 2 }\ngtr 1$$

nipunmalhotra93 · Answer

Wait a min. Is the taylor series given? Or did you find it?

phoenixfire · Answer

I found it, and it's known. Obviously I've ignored the $\large \frac{2}{\sqrt{\pi}}$ that is multiplied by all of it. http://en.wikipedia.org/wiki/Error_function#Taylor_series

kainui · Answer

@nipunmalhotra93  Actually it's not too hard to find the taylor series of erf, just plug in (-x^2) for the power series of e^x and then integrate it, that's why you have the (2n+1) in the power and divided by, just the power rule.

phoenixfire · Answer

I guess if I apply a different property for increasing
if $a_n=f(n),\forall n \in \mathbb{N}$ and $f'(x) \gt 0, \forall x\in [1,\infty)$
since $f'(x)=\frac{2}{\sqrt{\pi }}e^{-x^2}$ then as $x \rightarrow \infty$, $f'(x) \rightarrow 0$ but never reaching 0 it'll still satisfy the property stated above, and therefore be strictly increasing?
Is this a valid argument?

kainui · Answer

I think you're right, that looks like the way to go since you don't have to worry about infinity; infinity isn't part of your domain.

phoenixfire · Answer

@Kainui Alright, I'll just stick with that way. And your suggestion looks like it works, it gets rid of the pesky -1 and the x term ends up to the power of 4, so it's all positive and then when multiplied by the ignored factor $\large \frac{2}{\sqrt{\pi}}$ it's all greater than 1. 
But I don't like that logic (and too much algebra), sorry lol. I'll go with the derivative method.

kainui · Answer

Haha yeah fair enough. I honestly don't see why it shouldn't be true. After all, if $$\frac{a_{n+1}}{a_n}>1$$ then obviously since n is arbitrary, $$\frac{a_{n+2}}{a_{n+1}}>1$$ Just rearranging the inequalities: $$a_{n+2}>a_{n+1}$$$$a_{n+1}>a_{n}$$then$$a_{n+2}>a_{n}$$
$$\frac{a_{n+2}}{a_{n}}>1$$
It seems to be alright.

phoenixfire · Answer

But by the same logic of rearranging the inequalities $$\frac{a_{n+1}}{a_n}>1\Rightarrow \frac{a_n}{a_{n+1}}<1$$ which is true as it's just the reciprocal of my first attempt which is negative...

kainui · Answer

Nah, that's not right because dividing by a negative in an inequality reverses the direction of it, so it will still be ok.

kainui · Answer

You've just abstracted it away into a variable. Plus, it's false as it is, since no negative number is larger than 1. Well maybe. Now I am either confusing you or myself. Whatever. I'm done causing problems now lol.

phoenixfire · Answer

Hahaha, I'll just leave it as it. Thanks for your help!