# Convergence of $\sum_{n=1}^{\infty}\frac{1}{(3n+8)!}$

Determine whether or not this series converges or diverges $$\sum_{n=1}^{\infty}\dfrac{1}{(3n+8)!}$$

My attempt: I used the ratio test and ended up having $$\lim_{n\to\infty}\frac{1}{(3n+11)(3n+10)(3n+9)(3n+8)}$$ which I then plugged in $\infty$ for $n$ and got $1/\infty$ which makes the limit $1$. Since the limit is less than $1$ the series converges according to the ratio test. Is this correct?

• A related question. Jul 7, 2015 at 18:10

We can do more than prove convergence, we can compute it.

Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. Then: $$\sum_{n\geq 0}\frac{x^{3n}}{(3n)!} = \frac{1}{3}\left(e^{x}+e^{\omega x}+e^{\omega^2 x}\right)\tag{1}$$ hence, by differentiating both sides: $$\sum_{n\geq 0}\frac{x^{3n+2}}{(3n+2)!}= \frac{1}{3}\left(e^{x}+\omega\, e^{\omega x}+\omega^2\,e^{\omega^2 x}\right)\tag{2}$$ so, by evaluating at $x=1$: $$\sum_{n\geq 0}\frac{1}{(3n+2)!} = \frac{e}{3}-\frac{2}{3\sqrt{e}}\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)\tag{3}$$ then subtracting the first terms of the LHS: $$\sum_{n\geq 1}\frac{1}{(3n+8)!} = \color{red}{-\frac{20497}{40320}+\frac{e}{3}-\frac{2}{3\sqrt{e}}\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)}.\tag{4}$$

• Indeed a very nice solution. But I did not understand, how did you get the first line? Jul 8, 2015 at 7:15
• @Nilan: from the discrete Fourier transform: $1^n+\omega^n+\omega^{2n}$ equals $3$ if $3\mid n$ and zero otherwise, so to consider $\frac{1}{3}\left(f(x)+f(\omega x)+f(\omega^2 x)\right)$ is equivalent to take every monomial of the Taylor series of $f$ in which the exponent is a multiple of three. Jul 8, 2015 at 8:58
• Thank you for your valuable comment. I'll study this myself. +1 Jul 8, 2015 at 9:51

$$\lim_{n\to\infty}\frac{1}{(3n+11)(3n+10)(3n+9)} = 0 < 1$$

The limit of the above is $0$ since informally, as you've said $1/\infty = 0$. Take any small number and divide it by increasingly huger numbers on your calculator, you'll see the result approaching $0$. Since obviously $0 < 1$, the series converges by the ratio test.

Also, as Jack pointed out below, the $3n+8$ should not appear in the denominator, since $$\frac{(3n+11)!}{(3n+8)!}=(3n+11)(3n+10)(3n+9)$$

• The $(3n+8)$ should not be there, $\frac{(3n+11)!}{(3n+8)!}=(3n+11)(3n+10)(3n+9)$. Anyway, that does not change much. Jul 7, 2015 at 16:49
• @JackD'Aurizio Woopsies, thanks for pointing that out, I edited my answer and credited you. :-) Jul 7, 2015 at 16:54

You are right.

You may also notice that $$(3n+8)!>n^2,\qquad n\geq1,$$ giving $$0<\sum_1^{\infty}\frac1{(3n+8)!}<\sum_1^{\infty}\frac1{n^2}<+\infty$$ and thus the convergence of your initial series by the comparison test.

Yet another way to do this one. See that $$\sum \frac{1}{n!}$$ is convergent isn't hard. Then note that $$\frac{1}{(3n+2)!} < \frac{1}{n!}$$ and use the Comparison Test.