Proving the inequality $\tan(1)\le\sum_{k=1}^{\infty} \frac{\sin(1/k^2)}{\cos^2 (1/(k+1))}$

If $k \in \mathbb{N}$, $k \geq 2$ and $x \in [0, \sqrt{\frac{1}{2}}]$ then

$$\frac{\partial}{\partial x}\sin(\frac{x}{k}) \cos(x) = \frac{1}{k} \cos(\frac{x}{k}) \cos(x) - \sin(\frac{x}{k}) \sin(x) \geq \frac{1}{2 k} - \frac{x^2}{k} \geq 0 $$

so $\sin(\frac{x}{k})\cos(x)$ is non-decreasing on this interval. In particular

$$ \frac{\sin(\frac{1}{k(k+1)})}{\cos(\frac{1}{k})} \leq \frac{\sin(\frac{1}{k^2})}{\cos(\frac{1}{k+1})} $$

for all $k\geq 2$. This inequality also holds in fact for $k=1$ since $\sin(1) < \sin(\pi - 2) = \sin(2)$.

Now we get the following inequalities for all $k \geq 1$: $$\begin{eqnarray} \tan(\frac{1}{k}) - \tan(\frac{1}{k+1}) &=& \frac{\sin(\frac{1}{k}) \cos(\frac{1}{k+1}) - \cos(\frac{1}{k}) \sin(\frac{1}{k+1})}{\cos(\frac{1}{k}) \cos(\frac{1}{k+1})}\\ & = &\frac{\sin(\frac{1}{k(k+1)})}{\cos(\frac{1}{k}) \cos(\frac{1}{k+1})} \leq \frac{\sin(\frac{1}{k^2})}{\cos^2(\frac{1}{k+1})} \end{eqnarray} $$

and summing from $1$ to $\infty$ gives the required result.