We set $C = \exp\left(\frac{I}{2\pi}\right)$ with $I = 4\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}$. The calculator gives:
$$\begin{aligned}
\exp\left(\frac{2}{\pi}\sum_{k=0}^5 \frac{(-1)^k}{(2k+1)^2}\right) &\approx 1{,}78774486868 \\
\exp\left(\frac{2}{\pi}\sum_{k=0}^6 \frac{(-1)^k}{(2k+1)^2}\right) &\approx 1{,}79449196958
\end{aligned}$$
Can we deduce the rounding of $C$ to $10^{-2}$ precision? If yes, give the value of this rounding. In any case, justify the answer properly.