I think I may have a cool approach to this.
\(\displaystyle \L\\\sum_{n=0}^{\infty}\frac{(-3)^{n}}{(2n)!}\)
Use the Taylor expansion for cos(x).
\(\displaystyle \L\\cos(x)=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n)!}\)
Let's go to 5 terms:
cos(x)=\(\displaystyle \L\\1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+\frac{x^{8}}{40320}-............\)
Now, expand your series.
\(\displaystyle \L\\1-\frac{3}{2}+\frac{9}{24}-\frac{27}{720}+\frac{81}{40320}-..................\)
The numerators are powers of 3.
Therefore, \(\displaystyle x=\sqrt{3}\) in the series.
Therefore, hence, and whereby it converges to \(\displaystyle \H\\cos(\sqrt{3})\)
