# Some Simple Lemmas For Bounding Power Series (Part 2)

Often functions are defined or approximated by infinitely long sums which can be thought of as polynomials with an infinite degree. For example,
$$\cos(x) = 1- x^2/2! + x^4/4! +x^6/6! + ….$$
(where $n! = n \cdot (n-1) \cdot (n-2) \cdots \cdot1$ and $4!=4\cdot 3\cdot 2\cdot 1$.) These sums are usually referred to as infinite series, and you can often use the first few terms to approximate the sum of all of the terms. For example,
$$\cos(x) \approx 1 – x^2/2$$when $|x| < 0.2$

In my last post, I presented some bounds on the error that occurs when you approximate an alternating series with the first few terms of the series. An alternating series is a infinite sum where the sign of the terms alternate between positive and negative. In this post, I will present a fairly simple bound on the error of approximating an infinite series where the terms grow no faster than exponentially.

We will use this well known, useful fact (#87 in the “100 Most useful Theorems and Ideas in Mathematics“).

Fact 1: If $|x|<1$,

$$1/(1-x) = 1 +x +x^2 + \cdots.$$

Proof: Fix any $x\in(-1,1)$. Let $S_n = 1+x+\cdots+x^n.$ Then
$\lim_{n\rightarrow\infty} S_n$ exists by Ratio test, and
\begin{aligned} S_n (1-x) &= 1- x^{n+1} \\ \lim_{n\rightarrow\infty} S_n (1-x)&= \lim_{n\rightarrow\infty} (1- x^{n+1})\\ (1-x)\lim_{n\rightarrow\infty} S_n&= 1\\ \lim_{n\rightarrow\infty} S_n&= 1/(1-x).\quad\mathrm{□} \end{aligned}

Lemma: Suppose $a_1, a_2, \ldots$ is a sequence of real numbers, $n$ is a positive integer, and $\alpha$ is a real number such that:

1. $0\leq \alpha < 1,$ and
2. $\alpha |a_i| \geq |a_{i+1}|$ for all integers $i\geq n+1$.

Then $S = \sum_{a=1}^\infty a_i$ exists, and $$\left| S – \sum_{i=1}^{n} a_i \right| \leq |a_{n+1}|/(1-\alpha).$$

Proof: $S$ exists by Ratio Test.  Now, I claim that for all positive integers $j$, $$|a_{n+j}| \leq \alpha\,^{j-1} |a_{n+1}|.$$ This holds almost trivially when $j=1$. And if we assume that the claim holds for $j=k$, then

$$|a_{n+k+1}| \leq \alpha |a_{n+k}| \leq \alpha\; \alpha\,^{k-1} |a_{n+1} | = \alpha\,^k |a_{n+1}|$$ which proves that the claim holds for $j=k+1$, which in turn proves that the claim is true by induction.

Now by the claim and Fact 1, \begin{aligned} \left| S – \sum_{i=1}^{n} a_i \right| &= \left|\sum_{i=n+1}^\infty a_i\right|\\ &\leq \sum_{i=n+1}^\infty | a_i | \\ &\leq \sum_{i=0}^\infty \alpha\,^i |a_{n+1}| \\ &= |a_{n+1}|/ (1-\alpha) \end{aligned}which proves the lemma.$\quad\mathrm{□}$

### Examples

Example 1: By definition,

$$\exp(x) = 1 + x + x^2/2! + x^3/3! + \cdots.$$
If $|x| <1$, then the lemma holds for $n=2$ and $\alpha=1/3$, so
$$| \exp(x) – (1+x) | \leq |x^2/2|/(1-\alpha) = \frac34 x^2.$$
We can also apply the lemma for $|x|<1$, $n=3$, and $\alpha = 1/4$ to get
$$| \exp(x) – (1+x+x^2/2) | \leq |x^3/6|/(1-\alpha) = \frac29 |x^3|.$$

$\mathrm{ }$

Example 2: By Fact 1 and integration,
$$-\log(1-x) = x + x^2/2+x^3/3+ x^4/4+\cdots$$when $|x|<1$. We can apply the lemma for $|x|<1$, $n=1,2,$ or 3, and $\alpha =|x|$ to get
\begin{aligned} |-\log(1-x) – x | & \leq |x^2/2|/(1-|x|),\\ |-\log(1-x) – (x+x^2/2) | &\leq |x^3/3|/(1-|x|),\mathrm{\ and}\\ |-\log(1-x) – (x+x^2/2+x^3/3) | &\leq |x^4/4|/(1-|x|).\\ \end{aligned}