# The Unapologetic Mathematician

## Infinite Series

And now we come to one of the most hated parts of second-semester calculus: infinite series. An infinite series is just the sum of a (countably) infinite number of terms, and we usually collect those terms together as the image of a sequence. That is, given a sequence $a_k$ of real numbers, we define the sequence of “partial sums”:

$\displaystyle s_n=\sum\limits_{k=0}^na_k$

and then define the sum of the series as the limit of this sequence:

$\displaystyle\sum\limits_{k=0}^\infty a_k=\lim\limits_{n\rightarrow\infty}s_n$

Notice, though, that we’ve seen a way to get finite sums before: using step functions as integrators. So let’s use the step function $\lfloor x\rfloor$, which is defined for any real number $x$ as the largest integer less than or equal to $x$.

This function has jumps of unit size at each integer, and is continuous from the right at the jumps. Further, over any finite interval, its total variation is finite. Thus if $f$ is any function continuous from the left at every integer it will be integrable with respect to $\lfloor x\rfloor$ over any finite interval. Further, we can easily see

$\displaystyle\int\limits_a^bf(x)d\lfloor x\rfloor=\sum\limits_{\substack{k\in\mathbb{Z}\\a

Now given any sequence $a_n$ we can define a function $f$ by setting $f(x)=a_{\lceil x\rceil}$ for any $x>-1$. That is, we round each number up to the nearest integer $n$ and then give the value $a_n$. This gives us a step function with the value $a_n$ on the subinterval $\left(n-1,n\right]$, which we see is continuous from the left at each jump. Thus we can always define the integral

$\displaystyle\int\limits_{-\frac{1}{2}}^bf(x)d\lfloor x\rfloor=\sum\limits_{k=0}^{\lfloor b\rfloor}a_k=s_{\lfloor b\rfloor}$

Then as we let $b$ go to infinity, $\lfloor b\rfloor$ goes to infinity with it. Thus the sum of the series is the same as the improper integral.

So this shows that any infinite series can be thought of as a Riemann-Stieltjes integral of an appropriate function. Of course, in many cases the terms $a_k$ of the sequence are already given as values $f(k)$ of some function, and in that case we can just use that function instead of this step-function we’ve cobbled together.

April 24, 2008 - Posted by | Analysis, Calculus

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