Tools for finding bounds on power series

97 Views Asked by At

Suppose someone hands you a series where the terms are some function of x and your goal is to find some bounds for the series for a given set of x-values (I'm thinking of power series in particular). What tools do you use to find these bounds? In particular, could you tell whether a power series in x is unbounded as x goes to infinity? I lack formal training in analysis, but please do not restrain your answers because of that.


For example, consider $$f(x) = \sum_{n=0}^\infty \frac{(-x)^n}{n!}$$ and for the moment do your best to forget that this is $\exp(-x)$. Let's say we looked at x = 100 and looked at the first terms of the series, we'd see some terms of rather large magnitude: $$f(100) = 1-100+5000-500000/3+... $$

The series is absolutely convergent, and the magnitude of terms eventually decreases. But how would one quickly tell that these large magnitude terms - after well over 100 such terms - would so perfectly cancel to leave a number less than 1? In the language of my question at top, how could one tell that $0 \leq f(x) \leq 1$ for $x \geq 0$?

Here's my attempt to show this particular series is between 0 and 1 for $x\geq0$: Consider $$g(x) = \sum_{m=0}^\infty \frac{(x)^m}{m!}$$ Note that $g(x)$ is unbounded as x grows larger, and in particular that $g(x)\geq 1$ for $x\geq 0$. Noting $$g(x) f(x) = \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^n x^{n+m}}{n!m!},$$ we see that on changing indices to $q=n+m$ we have $$g(x) f(x) = \sum_{q=0}^\infty \sum_{n=0}^q \frac{(-1)^n x^{q}}{n!(q-n)!}.$$ Thus $$g(x) f(x) = \sum_{q=0}^\infty \frac{x^q}{q!} \sum_{n=0}^q (-1)^n \binom{q}{n},$$ which simplifies to $$g(x) f(x) = \sum_{q=0}^\infty \frac{x^q}{q!} \delta_{q 0} = 1.$$

Thus $$f(x) = \frac{1}{g(x)},$$ and since $g(x)\geq 1$ for $x\geq 0$, we see that $$0 \leq f(x) \leq 1 $$ for $x\geq 0$.

This attempt to put bounds on f(x) for $x \in [0,\infty) $ is to demonstrate my thought process on putting bounds on power series (the technique I used rested solely on foreknowledge), and I look forward to reading about the tools you use to put bounds on more general power series.

1

There are 1 best solutions below

1
On

I'll open with one of the main analytical tools used, the comparison test. This is particularly useful for answering the 'bounded or unbounded' question.

Suppose $A = \sum_{i=1}^{\infty}a_{n}$ and $B = \sum_{i=1}^{\infty}b_{n}$ are series. If $a_{n} \leq b_{n}$ for all $n \in \mathbb{N}$ and we know that B is bounded, then A is bounded. Similarly, if we knew that A was unbounded, we would know that B must be unbounded.

For example, if we begin with the knowledge that $\sum_{i=1}^{\infty}\frac{1}{n}$ is unbounded, we can say that $\sum_{i=1}^{\infty}\frac{3}{n}$ is unbounded, as each term is greater than a known unbounded series. Similarly since $\sum_{i=1}^{\infty}\frac{1}{n^{2}}$ is bounded, we know that $\sum_{i=1}^{\infty}\frac{1}{4n^{2}}$ is bounded, as each term is smaller than that of the former series.

It's been a long time since I've done much analysis, there are a great many tools that other posters can hopefully explain better but I hope that's an interesting idea to get you started.