Math for Non-Geeks/Geometric series

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Geometric series are series of the form $\sum _{k=0}^{\mathrm{\infty }}{q}^k$. They are important within several proofs in real analysis. In particular, they are crucial for proving convergence or divergence of other series. We will derive some criteria using them, e.g. the ratio or the root criterion.

Geometric sum formula

File:Geometrische Reihe - Quatematik.webm We recall the geometric sum formula for partial sums of the geometric series. Template:Noprint The proof of the sum formula reads as follows:

Geometric series

File:Geometrische Reihe (Mathe-Song) – DorFuchs.webm

The geometric series $\sum _{k=0}^{\mathrm{\infty }}{r}^k$ converges for $r=\frac{1}{2}$, $r=\frac{1}{3}$ or $r=\frac{1}{4}$ .

We consider two cases: $|q|<1$ and $|q|\ge 1$.

Case $|q|<1$

We consider the geometric series $\sum _{k=0}^{\mathrm{\infty }}{q}^k$ for any $|q|<1$ , which especially means $q\ne 1$. The sum formula above applies to the partial sums in that case:

So the geometric series converges if and only if the sequence of partial sums ${\left(\frac{1-q^{n+1}}{1-q}\right)}_{n\in \mathbb{N}}$ converges. This is the case if and only if ${\left(q^n\right)}_{n\in \mathbb{N}}$ converges. We know that ${\left(q^n\right)}_{n\in \mathbb{N}}$ converges to $0$ if and only if $|q|<1$ and it converges to $1$ , if and only if $q=1$. In this section, we only care about the first case of convergence:

Now, let us determine its limit:

Case $|q|\ge 1$

For $|q|\ge 1$ , we have for all $k\in {\mathbb{N}}_0$, that $\left|q^k\right|\ge 1$. Therefore, the sequence ${\left(q^k\right)}_{k\in {\mathbb{N}}_0}$ cannot converge to 0. So teh series $\sum _{k=0}^{\mathrm{\infty }}{q}^k$ must diverge (this argument is called term test and will be considered in detail, later)

The divergence becomes particularly obvious, if $q$ is positive, e.g. for $q\ge 1$. In this case, for all $k\in \mathbb{N}$, we have $q^k\ge 1$ and may estimate the partial sums: $\sum _{k=1}^n{q}^k\ge \sum _{k=1}^{n}1=n$ So the sequence of partial sums is bounded from below by the sequence $(n{)}_{n\in \mathbb{N}}$ , which in turn diverges to $+\mathrm{\infty }$. So the series $\sum _{k=0}^{\mathrm{\infty }}{q}^k$ must diverge, as well.

Conclusion

We have learned: for $|q|>1$, $q=-1$ and $q=1$ , the geometric series diverges. These three cases can be concluded into one case $|q|\ge 1$ . However, if $|q|<1$ , then the geometric series converges to $\frac{1}{1-q}$:

Example problems

Problem 4

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