Question

The sum $\sum_{n=1}^{\infty} \frac{9n^2}{(3n)!}$ is equal to

• A. e + \frac{1}{\sqrt{e}} cos(\frac{\sqrt{3}}{2})
• B. \frac{2}{3}e - \frac{1}{\sqrt{e}} cos(\frac{\sqrt{3}}{2})
• C. \frac{2}{3}(e + \frac{1}{\sqrt{e}} cos(\frac{\sqrt{3}}{2}))
• D. \frac{2}{3}(e - \frac{1}{\sqrt{e}} cos(\frac{\sqrt{3}}{2}))

Answer 1

Answer: (B)

\frac{2}{3}e - \frac{1}{\sqrt{e}} cos(\frac{\sqrt{3}}{2})

Answer 2

The sum is $S = \sum_{n=1}^{\infty} \frac{9n^2}{(3n)!}$. Using the identity $\frac{k^2}{k!} = \frac{1}{(k-2)!} + \frac{1}{(k-1)!}$ for $k \ge 2$, and setting $k=3n$, we get $\frac{(3n)^2}{(3n)!} = \frac{1}{(3n-2)!} + \frac{1}{(3n-1)!}$ for $n \ge 1$. Thus, $S = \sum_{n=1}^{\infty} \left(\frac{1}{(3n-2)!} + \frac{1}{(3n-1)!}\right)$. This can be split into two sums: $S_A = \sum_{n=1}^{\infty} \frac{1}{(3n-2)!} = \frac{1}{1!} + \frac{1}{4!} + \frac{1}{7!} + \dots$ and $S_B = \sum_{n=1}^{\infty} \frac{1}{(3n-1)!} = \frac{1}{2!} + \frac{1}{5!} + \frac{1}{8!} + \dots$. These sums correspond to the Taylor series expansions of $e^x$ evaluated at $x=1$ using complex cube roots of unity. Let $S_0(x) = \sum_{m=0}^{\infty} \frac{x^{3m}}{(3m)!}$, $S_1(x) = \sum_{m=0}^{\infty} \frac{x^{3m+1}}{(3m+1)!}$, $S_2(x) = \sum_{m=0}^{\infty} \frac{x^{3m+2}}{(3m+2)!}$. At $x=1$: $3S_0(1) = e + e^\omega + e^{\omega^2}$ $3S_1(1) = e + \omega^2 e^\omega + \omega e^{\omega^2}$ $3S_2(1) = e + \omega e^\omega + \omega^2 e^{\omega^2}$ The sum $S$ is $S_A + S_B = S_1(1) + S_2(1)$. $3S = 3(S_1(1) + S_2(1)) = (e + \omega^2 e^\omega + \omega e^{\omega^2}) + (e + \omega e^\omega + \omega^2 e^{\omega^2})$ $3S = 2e + (\omega^2+\omega)e^\omega + (\omega+\omega^2)e^{\omega^2}$ Since $\omega^2+\omega = -1$, we have $3S = 2e - e^\omega - e^{\omega^2}$. Using $e^\omega + e^{\omega^2} = 2e^{-1/2}\cos(\frac{\sqrt{3}}{2})$, we get: $3S = 2e - 2e^{-1/2}\cos(\frac{\sqrt{3}}{2})$ $S = \frac{2}{3}e - \frac{2}{3}e^{-1/2}\cos(\frac{\sqrt{3}}{2})$.