1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ A related question. $\endgroup$ – Lucian Jul 7 '15 at 18:10
3
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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 '15 at 16:49
  • $\begingroup$ @JackD'Aurizio Woopsies, thanks for pointing that out, I edited my answer and credited you. :-) $\endgroup$ – Zain Patel Jul 7 '15 at 16:54
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Indeed a very nice solution. But I did not understand, how did you get the first line? $\endgroup$ – Bumblebee Jul 8 '15 at 7:15
  • 1
    $\begingroup$ @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 '15 at 8:58
  • 1
    $\begingroup$ Thank you for your valuable comment. I'll study this myself. +1 $\endgroup$ – Bumblebee Jul 8 '15 at 9:51
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.