# The Unapologetic Mathematician

## Examples of Convergent Series

Today I want to give two examples of convergent series that turn out to be extremely useful for comparisons.

First we have the geometric series whose terms are the sequence $a_n=a_0r^n$ for some constant ratio $r$. The sequence of partial sums is

$\displaystyle\sum\limits_{k=0}^na_0r^k=a_0\left(1+r+r^2+...+r^n\right)$

If $r\neq1$ we can multiply this sum by $\frac{1-r}{1-r}$ to find

$\displaystyle\sum\limits_{k=0}^na_0r^k=a_0\frac{1-r^{n+1}}{1-r}$

Then as $n$ goes to infinity, this sequence either blows up (for $|r|>1$) or converges to $\frac{a_0}{1-r}$ (for $|r|<1$). In the border case $r=\pm1$ we can also see that the sequence of partial sums fails to converge. Thus the geometric series converges if and only if $|r|<1$, and we have a nice simple formula telling us the sum.

The other one I want to hit is the so-called $p$-series, whose terms are $a_n=n^{-p}$ starting at $n=1$. Here we use the integral test to see that

$\displaystyle\lim\limits_{n\rightarrow\infty}\left(\sum\limits_{k=1}^n\frac{1}{n^p}-\int\limits_1^n\frac{dx}{x^p}\right)=D$

so the sum and integral either converge or diverge together. If $p\neq1$ the integral gives $\frac{n^{1-p}-1}{1-p}$, which converges for $p>1$ and diverges for $p<1$.

If $p=1$ we get $\ln(n)$, which diverges. In this case, though, we have a special name for the limit of the difference $D$. We call it “Euler’s constant”, and denote it by $\gamma$. That is, we can write

$\displaystyle\sum\limits_{k=1}^n\frac{1}{k}=\ln(n)+\gamma+e(n)$

where $e(n)$ is an error term whose magnitude is bounded by $\frac{1}{n}$.

In general we have no good value for the sums of these series, even where they converge. It takes a bit of doing to find $\sum\frac{1}{n^2}=\frac{\pi^2}{6}$, as Euler did in 1735 (solving the “Basel Problem” that had stood for almost a century), and now we have values for other even natural number values of $p$. The sum $\sum\frac{1}{n^3}$ is known as Apéry’s constant, after Roger Apéry who showed that it was irrational in 1979. Yes, we didn’t even know whether it was a rational number or not until 30 years ago. We have basically nothing about odd integer values of $p$.

If we say $s$ instead of $p$, and let $s$ take complex values (no, I haven’t talked about complex numbers yet, but some of you know what they are) we get Riemann’s function $\zeta(s)=\sum\frac{1}{n^s}$, which is connected to some of the deepest outstanding questions in mathematics today.

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Posted by John Armstrong | Analysis, Calculus

## 4 Comments	»

1. [...] if we set , this tells us that . Then the comparison test with the geometric series tells us that [...]

Pingback by | May 5, 2008 | Reply

2. [...] the complex norm is multiplicative, everything for the geometric series goes through again: if , and it diverges if . The case where is more complicated, but it can be [...]

Pingback by | August 28, 2008 | Reply

3. [...] the final summation converges because it’s a geometric series with initial term and ratio . This implies that [...]

Pingback by | December 14, 2009 | Reply

4. [...] this is a chunk of a geometric series; since , the series must converge, and so we can make this sum as small as we please by choosing [...]

Pingback by | May 6, 2011 | Reply

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