A few days ago, I wrote several proofs that $\sin(2°)/2 > \sin(3°)/3$. One of those proofs involves power series and two simple, useful lemmas.

Leibniz bound on alternating series (1682)

Theorem: Suppose $a_1, a_2, \ldots$ is a sequence of real numbers such that:

1) $|a_i| \geq |a_{i+1}|$ for positive all integers $i$,

2) $a_1 >0$,

3) $a_j \cdot a_{j+1} <0$ for positive integers $j$ (i.e. the sequence alternates in sign), and

4) $\sum_{a=1}^\infty a_i$ exists.

Then $$0 \leq \sum_{i=1}^\infty a_i \leq a_1.$$

(See Theorem 2 in this paper.) The following two lemmas are almost the same and almost follow from the Leibniz bound. (If you make the strict inequalities in the Lemmas non-strict, then you can prove them in one sentence using the Leibniz theorem.)

Lemma 1: Suppose $a_1, a_2, \ldots$ is a sequence of real numbers such and $n$ is a positive integer such that:

1) $|a_i| \geq |a_{i+1}|$ for all integers $i\geq n$,

2) $a_n <0$,

3) $a_j \cdot a_{j+1} <0$ for all integers $j\geq n$ (i.e. the sequence alternates in sign), and

4) $\sum_{a=1}^\infty a_i$ exists.

Then $$\sum_{i=1}^\infty a_i > \sum_{i=1}^n a_i .$$

Lemma 2: Suppose $a_1, a_2, \ldots$ is a sequence of real numbers such and $n$ is a positive integer such that:

1) $|a_i| \geq |a_{i+1}|$ for all integers $i\geq n$,

2) $a_n >0$, and

3) $a_j \cdot a_{j+1} <0$ for all integers $j\geq n$ (i.e. the sequence alternates in sign), and

4) $\sum_{a=1}^\infty a_i$ exists.

Then $$\sum_{i=1}^\infty a_i < \sum_{i=1}^n a_i .$$

Proof: Here is a proof that does not use the Leibniz Theorem.

The hypotheses imply that $a_{n +2k +1}<0$, $a_{n +2k +2}>0$, and

$$ a_{n +2k +1} + a_{n +2k +2}<0$$

for all integers $k\geq 0$. So,

$$\begin{aligned}

\sum_{i=1}^\infty a_i &= \sum_{i=1}^n a_i + \sum_{i=n+1}^\infty a_i \\

&= \sum_{i=1}^n a_i + \sum_{k=0}^\infty a_{n +2k +1} + a_{n +2k + 2j } \\

&< \sum_{i=1}^n a_i.\quad ■

\end{aligned}$$

Example 1: $$\sin(x) = \sum_{i=0}^\infty (-1)^i x^{2i+1}/{(2i+1)!},$$so $$x> \sin(x) > x-x^3/6$$when $0<x<\sqrt{3}$ because the absolute value of the terms are decreasing for that interval. (In reality, it is true for any real $x$, but you need more than the Lemmas above to prove that.) Furthermore, $$x-x^3/6 + x^5/120 > \sin(x) > x-x^3/6 + x^5/120 – x^7/5040$$when $0<x<\sqrt{20}$. (Once again, it is true for any real $x$.)

Example 2: $$\cos(x) = \sum_{i=0}^\infty (-1)^i x^{2i}/{(2i)!},$$so $$1> \cos(x) > 1-x^2/2$$when $0<|x|<\sqrt{2}$ because the absolute value of the terms are decreasing for that interval.

Example 3: $$\log(1+x) = \sum_{i=1}^\infty (-1)^{i+1} x^{i}/i$$when $|x|<1$, so $$x> \log(x+1) > x-x^2/2$$when $0<x<1$ because the absolute value of the terms are decreasing for that interval. Furthermore, $$x-x^2/2 + x^3/3 > \log(x+1) > x-x^2/2 + x^3/3-x^4/4$$when $0<x<1$ because the absolute value of the terms are decreasing for that interval.

Example 4: $$\mathrm{arctan}(x) = \sum_{i=0}^\infty (-1)^i x^{2i+1}/(2i+1),$$so $$x> \mathrm{arctan}(x) > x-x^3/3 $$when $0<x<1$ because the absolute value of the terms are decreasing for that interval. Furthermore, $$x-x^3/3+x^5/5 > \mathrm{arctan}(x) >x-x^3/3+x^5/5 -x^7/7$$when $0<x<1$ because the absolute value of the terms are decreasing for that interval.