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Evaluate

$$\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\pi^{6k}}{(6k)!}$$

I was trying to find a closed form for this sum

$$\sum_{k=1}^{\infty} (-1)^{k+1}\frac{x^{6k}}{(6k)!}$$

I believe there is something to do with $$\cos(x)=\sum_{k=0}^{\infty} (-1)^{k}\frac{x^{2k}}{(2k)!}$$

I do not have a clear idea where to started. I am wondering if someone would be able help me out !

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  • $\begingroup$ You've written down the series for sine, not cosine. $\endgroup$
    – Olivier Moschetta
    Commented Sep 19, 2016 at 19:08
  • $\begingroup$ Remember that $\left(a^b\right)^c = a^{bc}$. Therefore $\left(\pi^3\right)^{2k} = \pi^{6k}$. $\endgroup$
    – Matt
    Commented Sep 19, 2016 at 19:12
  • $\begingroup$ Maybe you're looking at a general expansion of $cos(x^3)$ $\endgroup$
    – Student
    Commented Sep 19, 2016 at 19:13
  • $\begingroup$ @ShreyAryan And maybe not, as writing down the expansion of $\cos(x^3)$ shows. $\endgroup$
    – Did
    Commented Sep 20, 2016 at 5:17

2 Answers 2

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Let $\omega=\exp\frac{2\pi i}{3}$. Notice that $1+\omega^n+\omega^{2n}$ equals $3$ if $3\mid n$ and zero if $3\nmid n$. Deduce that

$$ \sum_{k\geq 0}\frac{(-1)^k x^{6k}}{(6k)!}=\sum_{3\mid k}\frac{(-1)^k x^{2k}}{(2k)!}=\frac{\cos(x)+\cos(\omega x)+\cos(\omega^2 x)}{3}.\tag{1}$$ Evaluate both sides at $x=\pi $ to get $$ \sum_{k\geq 0}\frac{(-1)^k x^{6k}}{(6k)!}=-\frac{1}{3},\tag{2}$$ profit.

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  • $\begingroup$ very clear solution (+1) $\endgroup$
    – Refaat M. Sayed
    Commented Sep 19, 2016 at 19:40
  • $\begingroup$ Stupid question: What is the notation $3|k$ for? $\endgroup$
    – Enrico M.
    Commented Sep 19, 2016 at 19:41
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    $\begingroup$ @FourierTransform: $3\mid k$ means "$3$ divides $k$", i.e. the sum is carried on over all the $k\geq 0$ that are multiples of $3$. $\endgroup$
    – Jack D'Aurizio
    Commented Sep 19, 2016 at 19:43
  • $\begingroup$ @+1 and mini upvote and thank you!! $\endgroup$
    – Enrico M.
    Commented Sep 19, 2016 at 19:43
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    $\begingroup$ would you explain how did you come up with $1+\omega^n+\omega^{2n}$ equals $3$ $\endgroup$
    – ADAM
    Commented Sep 19, 2016 at 20:11
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This series is not a simple function. Indeed the general series for $x$ gives:

$$\sum_{k = 1}^{+\infty} (-1)^{k+1}\frac{x^{6k}}{(6k)!} = 1-\, _0F_5\left(\frac{1}{6},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{5}{6};-\frac{x^6}{46656}\right)$$

Namely is an HyperGeometric Function.

In you case, though, the sum is simply

$$\frac{4}{3}$$

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  • $\begingroup$ I do not agree. By the discrete Fourier transform, the LHS is a simple function. $\endgroup$
    – Jack D'Aurizio
    Commented Sep 19, 2016 at 19:36
  • $\begingroup$ @JackD'Aurizio I would be honoured if you showed me the whole process. It's since lots of time I wanted to see some application of the FFT! In any case, I remember that function (well not that one really, but a more general one) tabulated in a book of functions, and unless I misread, I remembered it was so! Correct me of course! $\endgroup$
    – Enrico M.
    Commented Sep 19, 2016 at 19:38
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    $\begingroup$ It is not the Fast Fourier Transform, it is the Discrete Fourier transform, whose principle entirely lies in the first line of my proof. The characteristic function of the integers $n\equiv 0\pmod{k}$ can be written in terms of a combination of $n$-th powers of $k$-th roots of unity. $\endgroup$
    – Jack D'Aurizio
    Commented Sep 19, 2016 at 19:40
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    $\begingroup$ en.wikipedia.org/wiki/Discrete_Fourier_transform $\endgroup$
    – Jack D'Aurizio
    Commented Sep 19, 2016 at 19:41
  • $\begingroup$ Sorry, I meant DFT of course! -.- Typo! $\endgroup$
    – Enrico M.
    Commented Sep 19, 2016 at 19:42

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