# Big-$O$ for $\frac{1}{1-x}$

I would like to show as $x\rightarrow 0$

$$\frac{1}{1-x}= 1+x^2+x^3+\dots+x^n +O(x^{n+1})$$

My inclination is to multiply by $1-x$ to get:

$1=(1-x)(1+x^2+\dots+x^n) +(1-x)O(x^{n+1})$

and then, for some constant $C$:

$1=x^{n+1}-1 +(1-x)C|x^n|$

But I am stuck here. I am a self-studier, but I tried to tag this as "homework" as it is that type of problem. Thanks.

You have

$$1+x+\cdots+x^n=\frac{1-x^{n+1}}{1-x}=\frac{1}{1-x}-\frac{x^{n+1}}{1-x}$$

Now, for $|x|<\frac12$,

$$\left|\frac{x^{n+1}}{1-x}\right|<2|x|^{n+1}$$

Hence

$$\frac{1}{1-x}=1+x+\cdots+x^n+O(x^{n+1})$$

You can prove the same on any interval $]-\varepsilon,\varepsilon[$ with $0<\varepsilon<1$, you will just get a factor $\frac{1}{1-\varepsilon}$ instead of $2$. You can't write the same inequality on $[-1,1]$ since the series is not convergent at $x=1$ or $x=-1$.

Just a side note: since $x^n=O(x^n)$, you could as well write $1+x+\cdots+x^{n-1}+O(x^n)$ in your question.

• Thanks, you are right, I had noticed my mistake in the consistency of the exponents; but should have fixed it as you suggest rather than as I did. Also please forgive me, while you answer is certainly excellent, the other responder came first, so I feel I should accept his answer. This aspect is always uncomfortable. With regards,
– user12802
Apr 9 '15 at 22:27
• @Andrew No worry :-) Apr 9 '15 at 22:29
– user12802
Apr 9 '15 at 22:29

One way to do this is to observe that the Taylor series for $\frac{1}{1-x}$ is:

$$\frac{1}{1-x} = \sum_{i = 0}^\infty x^i$$

Which converges for $\left \lvert x \right \rvert < 1$. So for any $n$ we can write:

$$\frac{1}{1-x} = 1 + x^2 + x^3 + \cdots + x^n + \sum_{i = n + 1}^\infty x^i$$

Now you simply need to prove that $\sum_{i = n + 1}^\infty x^i$ is $O(x^n)$ as $x \to 0$. I will show you a proof of that if you'd like, but since you are doing self-study it might be a good idea to do it yourself if you can.

• Thanks - I'll give it a try. Regards,
– user12802
Apr 9 '15 at 22:14