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Calculate the sum of the infinite series $$\sum_{n=0}^{\infty}\frac{1}{(3n)!}$$ by solving an aptly chosen differential equation.

I know that one can solve a differential equation by assuming that we can write the solution as a power series in the form $$y(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$$ and then find all the different values for $a_n$'s. I'm trying to figure out how to to it the other way around? How am I supposed to find the differential equation when I have the infinite sum already?

Update I've started off supposing there exists some solution, to the differential equation of the form $$p(x)y''(x)+q(x)y'(x)+r(x)y(x)=0,$$ that can be written as $$y(x)=\sum_{n=0}^\infty a_n(x-x_0)^n. $$ Since I already know this solution should be of the form $$y(x)=\sum_{n=0}^\infty \frac{1}{(3n)!}$$ I know that $a_n=\frac{1}{(3n)!}, x=1, x_0=0$. Furthermore, writing $$y'(x)=\sum_{n=1}^\infty n\ a_n(x-x_0)^{n-1} \ \ and \ \ y''(x)=\sum_{n=2}^\infty n(n-1)\ a_n(x-x_0)^{n-2}. $$ and filling this in the differential equation I finally end up with the equation $$\sum_{n=0}^\infty \left(p(x)\frac{(n+2)(n+1)}{(3(n+2))!} + q(x)\frac{(n+1)}{(3(n+1))!}+r(x)\frac{1}{(3n)!} \right)=0$$ I think that all the functions $p(x), q(x) \ $and $ r(x)$ should be evaluated at $x=1$. However, I don't know how to proceed.

Whizkid95
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2 Answers2

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In order to sum a series $\sum_{k=0}^\infty a_k$ of constant terms it often helps to consider the function $y(x):=\sum_{k=0}^\infty a_k x^k$. In the case at hand we have $$y(x)=\sum_{n=0}^\infty {x^{3n}\over (3n)!}=1+{x^3\over 3!}+{x^6\over 6!}+{x^9\over 9!}+\ldots\ .\tag{1}$$ Then $$y'(x)={3x^2\over3!}+{6x^5\over 6!}+{9x^8\over 9!}\ldots={x^2\over2!}+{x^5\over5!}+{x^8\over8!}+\ldots,\qquad y''(x)={x\over1!}+{x^4\over 4!}+{x^7\over 7!}+\ldots$$ and therefore $$y'''(x)=1+{x^3\over 3!}+{x^6\over 6!}+\ldots=y(x)\ .$$ It follows that this function satisfies the differential equation $$y'''-y=0\ .$$ Now find the general solution of this ODE. Among the $\infty^3$ solutions there is exactly one that also satisfies the initial conditions $y(0)=1$, $y'(0)=0$, $y''(0)=0$, as does the $y(x)$ in $(1)$. If $y_*(\cdot)$ is this solution then $y_*(1)$ is the sum of your series.

Christian Blatter
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  • Sorry for the late question, but why is $y'''-y=0$? I've been trying to calculate it, but I get $y'''= \sum_{n=0}^{\infty} \frac{1}{(3n+6)!}x^{3n+6}$. But $ \sum_{n=0}^{\infty} \frac{1}{(3n+6)!}x^{3n+6}- \sum_{n=0}^{\infty}\frac{1}{(3n)!}x^{3n}$ is not equal to $0$ – Fruitdruif Mar 23 '19 at 11:05
  • @Noah_K: Your $y'''$ is wrong. – Christian Blatter Mar 23 '19 at 11:40
  • Why do you write $3n$ as exponent for $x$? – The Coding Wombat Mar 23 '19 at 12:25
  • And how do you get $y(0)=1$? Do you assume that $0^0=1$? – The Coding Wombat Mar 23 '19 at 12:45
  • @TheCodingWombat If you write out the derivatives you'll see why he uses $3n$ as the exponent – Fruitdruif Mar 23 '19 at 12:52
  • @Noah_K I found $y'''(x)=\Sigma_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}=\Sigma_{n=0}^{\infty} \frac{x^{3(n-1)}}{(3(n-1))!}$. But this means the first term of the third derivative has a negative factorial in it. There does not seem to be consensus on how to deal with that. If a negative factorial is infinity, then indeed $y''' - y = 0$. But did the answerer choose the $3n$ exponent because that choice makes $y'''-y=0$ work; or did the $3n$ follow logically from the summation $\Sigma_{k=0}^{\infty} a_{k}x^{k}$? If he chose it to make the DE work, how to find this without knowing it in advance? – The Coding Wombat Mar 23 '19 at 13:30
  • The series will start at $n=1$ so you wont be dealing with a negative factorial. – esteban Mar 25 '19 at 13:22
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From a polynomial or power series $f(x)$ you can "punch out" the odd degree terms by taking the even part $\frac12(f(x)+f(-x))$. In a similar fashion you can produce "holes" with period $3$ in the coefficient sequence by combining $f(e^{ik\frac{2\pi}3}x)$, $k=-1,0,1$.

As the remaining coefficients in the given series can be interpreted to be every third coefficient of the exponential series, it can be written the evaluation at $x=1$ of $$ \frac13(e^x+e^{wx}+e^{\bar wx}) $$ with $w^3=1$, i.e., $w=e^{i\frac{2\pi}3}=\frac12(-1+i\sqrt3)$.

This can now easily be transformed into a linear ODE of order 3. (Which, arguably, is backwards from what is intended for the task. So you got to find an argument directly starting from the periodicity of $n!a_n$.)

Lutz Lehmann
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