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.

April 29, 2008 Posted by | Analysis, Calculus | 4 Comments

The Integral Test

Sorry for the delay. Students are panicking on the last day of classes and I have to write up a make-up exam for one who has a conflict at the scheduled time…

We can forge a direct connection between the sum of an infinite series and the improper integral of a function using the famed integral test for convergence.

I’ve spent a goodly amount of time last week trying to craft a proof hinging on converting the infinite sum to an improper integral using the integrator $\lfloor x\rfloor$, and comparing that one to those using the integrators $x$ and $x-1$. But it doesn’t seem to be working. If you can make a go of it, I’ll be glad to hear it. Instead, here’s a proof adapted from Apostol.

We let $f$ be a positive decreasing function defined on some ray. For our purposes, let’s let it be $\left[1,\infty\right)$, but we could use any other and adapt the proof accordingly. What we require in any case, though, is that the limit $\lim\limits_{x\rightarrow\infty}f(x)=0$. We define three sequences:

$\displaystyle s_n=\sum\limits_{k=1}^nf(k)$
$\displaystyle t_n=\int\limits_1^nf(x)dx$
$d_n=s_n-t_n$

First off, I assert that $d_n$ is nonincreasing, and sits between $f(n)$ and $f(1)$. That is, we have the inequalities

$0

To see this, first let’s write the integral defining $t_{n+1}$ as a sum of integrals over unit steps and notice that $f(k)$ gives an upper bound to the size of $f$ on the interval $\left[k,k+1\right]$. Thus we see:

$t_{n+1}\displaystyle\sum\limits_{k=1}^n\int\limits_k^{k+1}f(x)dx\leq\sum\limits_{k=1}^n\int\limits_k^{k+1}f(k)dx=\sum\limits_{k=1}^nf(k)=s_n$

From here we find that $f(n+1)=s_{n+1}-s_n\leq s_{n+1}-t_{n+1}=d_{n+1}$.

On the other hand, we see that $d_n-d_{n+1}=t_{n+1}-t_n-(s_{n+1}-s_n)$. Reusing some pieces from before, we see that this is

$\displaystyle\int\limits_n^{n+1}f(x)dx-f(n+1)\geq\int\limits_n^{n+1}f(n+1)dx-f(n+1)=0$

which verifies that the sequence $d_n$ is decreasing. And it’s easy to check that $d_1=f(1)$, which completes our verification of these inequalities.

Now $d_n$ is a monotonically decreasing sequence, which is bounded below by ${0}$, and so it must converge to some finite limit $D$. This $D$ is the difference between the sum of the infinite series and the improper integral. Thus if either the sum or the integral converges, then the other one must as well.

We can actually do a little better, even, than simply showing that the sum and integral either both converge or both diverge. We can get some control on how fast the sequence $d_n$ converges to $D$. Specifically, we have the inequalities $0\leq d_k-D\leq f(k)$, so the difference converges as fast as the function goes to zero.

To get here, we look back at the difference of two terms in the sequence:

$\displaystyle0\leq d_n-d_{n+1}\leq\int\limits_n^{n+1}f(n)dx-f(n+1)=f(n)-f(n+1)$

So take this inequality for $n=k$ and add it to that for $n=k+1$. We see then that $0\leq d_k-d_{k+2}\leq f(k)-f(k+2)$. Then add the inequality for $n=k+2$, and so on. At each step we find $0\leq d_k-d_{k+l}\leq f(k)-f(k+l)$. So as $l$ goes to infinity, we get the asserted inequalities.

April 29, 2008 Posted by | Analysis, Calculus | 1 Comment