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:

- $0\leq \alpha < 1,$ and
- $\alpha |a_i| \geq |a_{i+1}|$ 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}

$$