Proof for convergence of recursive series: $a_1 = \frac\pi4$, and $a_n =\frac12 \tan(a_{n-1})$

Question

If I define a sequence $a_n$: $$ a_1 = \frac{\pi}{4},~~ a_n =\frac{1}{2} \tan(a_{n-1}) $$ and then define a series: $$ \sum_{i=1}^n a_n $$

I can see why it converges, by the comparison test for $n\gt 6$
$a_n < \dfrac{1}{n^2}$ and we know $\dfrac{1}{n^2}$ converges as a p-series so
$a_n$ must converge.

But how do I formally prove it?

Answer 1

Since $\tan''(x)=2\sec^2(x)\tan(x)\ge0$ for $0\le x\le\frac\pi4$, $\tan(x)$ is convex for $0\le x\le\frac\pi4$. Therefore, for $0\le x\le\frac\pi4$, $$ \tan(x)\le\frac4\pi x\tag1 $$ Inductively, we get $$ a_n\le\frac\pi4\left(\frac2\pi\right)^{n-1}\tag2 $$ Thus, the series converges by comparison to a geometric series with ratio between $-1$ and $1$.

Answer 2

For sufficiently small positive angles,

$$\tan a<1.1\,a$$ and by induction the series is bounded by a converging geometric one (of common ratio $\dfrac{1.1}2$).