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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?

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  • $\begingroup$ A related question. $\endgroup$
    – Lucian
    Jul 7, 2015 at 18:10

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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: $$ \sum_{n\geq 1}\frac{1}{(3n+8)!} = \color{red}{-\frac{20497}{40320}+\frac{e}{3}-\frac{2}{3\sqrt{e}}\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)}.\tag{4}$$

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  • $\begingroup$ Indeed a very nice solution. But I did not understand, how did you get the first line? $\endgroup$
    – Bumblebee
    Jul 8, 2015 at 7:15
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    $\begingroup$ @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. $\endgroup$
    – Jack D'Aurizio
    Jul 8, 2015 at 8:58
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    $\begingroup$ Thank you for your valuable comment. I'll study this myself. +1 $\endgroup$
    – Bumblebee
    Jul 8, 2015 at 9:51
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$$\lim_{n\to\infty}\frac{1}{(3n+11)(3n+10)(3n+9)} = 0 < 1$$

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 $$\frac{(3n+11)!}{(3n+8)!}=(3n+11)(3n+10)(3n+9)$$

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

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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.

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