Hey guys, I've finished the first three parts, but I have no idea how to approach part d and part e. Any hints would help! Thanks!
Update: I've finished the first three parts. For part d, I found the taylor series for $sin(x)$. Since $p_k=ksin(\pi/k)$, I substitute x with $\pi/k$ and I get $p_k=\sum_{j=o}^\infty(-1)^jk(\pi/k)^{2j+1}/(2j+1)!$= $\sum_{j=o}^\infty(-1)^j(\pi)^{2j+1}k^{-2j}/(2j+1)!$. And I did the same with $P_k$. But still I don't know how to connect what I got with part d.
For (d): we will use the Taylor series for $\sin$ and $\tan$, namely $$ \sin x = \sum_{j=0}^\infty \frac{(-1)^j}{(2j+1)!}x^{2j+1}, \quad \tan x = \sum_{j=0}^\infty \frac{U_{2j+1}}{(2j+1)!}x^{2j+1} $$ for some constants $(U_j)_{j\in\mathbb{N}}$.
Detailed derivation:
We now have, for any $k\geq 4$, $$ \begin{align} p_k &= k \sin \frac{\pi}{k} = k \sum_{j=0}^\infty \frac{(-1)^j}{(2j+1)!} \left(\frac{\pi}{k}\right)^{2j+1} \\ &= \sum_{j=0}^\infty \frac{(-1)^j}{(2j+1)!} \frac{k\pi^{2j+1}}{k^{2j+1}} = \sum_{j=0}^\infty \frac{(-1)^j}{(2j+1)!} \frac{\pi^{2j+1}}{k^{2j}} \\ &= \frac{(-1)^0}{(2\cdot 0+1)!} \frac{\pi^{2\cdot 0+1}}{k^{2\cdot 0}} + \sum_{j=1}^\infty \frac{(-1)^j}{(2j+1)!} \frac{\pi^{2j+1}}{k^{2j}} \\ &= \pi + \sum_{j=1}^\infty \frac{(-1)^j}{(2j+1)!} \frac{\pi^{2j+1}}{k^{2j}} \\ &= \pi + \sum_{j=1}^\infty \frac{q_j}{k^{2j}} \end{align} $$ for $q_j \stackrel{\rm def}{=} \frac{(-1)^j\pi^{2j+1}}{(2j+1)!}$.
Similarly, $$ \begin{align} P_k &= k \tan \frac{\pi}{k} = k \sum_{j=0}^\infty \frac{U_{2j+1}}{(2j+1)!} \left(\frac{\pi}{k}\right)^{2j+1} \\ &= \pi + \sum_{j=1}^\infty \frac{U_{2j+1}}{(2j+1)!} \frac{\pi^{2j+1}}{k^{2j}} \\ &= \pi + \sum_{j=1}^\infty \frac{Q_j}{k^{2j}} \end{align} $$ for $Q_j \stackrel{\rm def}{=} \frac{U_{2j+1}\pi^{2j+1}}{(2j+1)!}$.