find exact values for the summation

[spdrmncoo]

i have a hard time to solve it

find exact values for the summation

sigma [ n = 0 to infinity ] ((-3)^n) / (2n)!

would be glad to have help

 

[galactus]

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 \(\displaystyle \H\\cos(\sqrt{3})\)