# Expressing a geometric series as the sum of the first n terms plus an “error term”?

Given a geometric series, how would you do this?

For example, how would this be done if the geometric series in question as is as follows?:

$$\frac{1}{(1 - (-x^2))}$$

-

## 2 Answers

$$\sum_{n=0}^\infty\, r^n \ = \ \sum_{n=0}^{m-1}\, r^n + \sum_{n=m }^\infty\, r^n \ = \ \underbrace{ 1-r^m \over 1-r}_{\text{first }m\text{ terms}} + \underbrace{ {r^{m }\over 1-r }}_{\text {error}}.$$

-
I'm confused.. what do r, m, and n represent? r is -x^2? – Josh1billion Dec 10 '11 at 2:48
@Josh1billion $r$ is the ratio, your $-x^2$. $n$ is the index counter and $m-1$ gives the final index (there will be $m$ total terms then). e.g $$\sum_{n=0}^5 r^n =r^0+r^1+r^2+r^4+r^5,$$ for $m=6$. – David Mitra Dec 10 '11 at 8:14

suppose that $a_n$ is a geometric series i.e. the sequence is: $$a_0 ,\quad a_1=a_0\times q,\quad a_2=a_1\times q = a_0 \times q^2,\quad \cdots , \quad a_n=a_0 \times q^n$$

The summation of $n$ elements of this sequence is: $$\sum_{i=0}^{n}a_i=a_0+a_1+a_2+\cdots+a_n=$$ $$a_0+a_0.q+a_0.q^2+\cdots+a_0.q^n=$$ $$a_0(1+q+q^2+\cdots+q^n) = a_0 \frac{q^{n+1}-1}{q-1}$$

for the proof of the last equation you can use Gauss's rule or simply divde $q^{n+1}-1$ to $q-1$.

when $-1<q<+1$ in the infinite sum ($n\rightarrow\infty$) $q^{n+1}\rightarrow 0$ (because it became smaller and smaller). So we have: $$\sum_{i=0}^{\infty} a_i= \frac{a_0(0-1)}{q-1}= \frac{-a_0}{q-1}= \frac{a_0}{1-q}$$

Similarly in the power series when $|x|<1$ we have: $$\sum_{i=0}^{\infty}x^i=\frac{1}{1-x}$$ and this our first formula in power series.

By replacing $x \rightarrow (-x)$ we conclude $$\sum_{i=0}^{\infty}(-x)^i=\sum_{i=0}^{\infty}(-1)^i.x^i=\frac{1}{1+x}$$

Now in the above formula we replace $x \rightarrow x^2$ and we conclude

$$\sum_{i=0}^{\infty}(-1)^i.x^{2i}=\frac{1}{1+x^2}$$

and this is what you want i.e.:

$$\frac{1}{1-(-x^2)}=\frac{1}{1+x^2}=1-x^2+x^4-x^6+\cdots$$

this is a geometric series with first element $a_0 = 1$ and common ratio $q=-x^2$.

-