Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?:

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

share|cite|improve this question
up vote 6 down vote accepted

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

share|cite|improve this answer
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


and this is what you want i.e.:


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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.