Find a closed form for this infinite sum: $ 1+\frac 1 2 +\frac{1 \times2}{2 \times 5}+\frac{1 \times2\times 3}{2 \times5\times 8}+ \dots$

Question

How to find a closed form for the expression??

$$ 1+\frac 1 2 +\frac{1 \times2}{2 \times 5}+\frac{1 \times2\times 3}{2 \times5\times 8}+\frac{1\times 2\times 3\times 4}{2 \times 5\times 8\times 11}+ \cdots$$

Wolfram alpha gives,

$$\frac{3}{2}+\frac{\ln(\sqrt[3]{2}-1)}{4\sqrt[3] {2}}+\frac{\sqrt{3}}{2\sqrt[3]{2}}\arctan\frac{\sqrt{3}}{2\sqrt[3]{2}-1}$$

Answer 1

First of all, let us note that $$\int_0^1(1-x)^{n-1}x^{-1/3}\,dx=\mathrm{B}\left(n,\frac23\right)=\frac{\Gamma(n)\Gamma(\frac23)}{\Gamma(n+\frac23)}=\frac{(n-1)!\,\Gamma(\frac23)}{(n-\frac13)(n-\frac43)\ldots\cdot \frac23 \Gamma(\frac23)}=\frac{3^n (n-1)!}{2\cdot 5\cdot\ldots\cdot(3n-1)},$$ which is almost the main term of our series (becomes it after multiplicating by $n/3^n$). Then the series in question without initial term $1$ equals to $$S=\sum_{n=1}^\infty \frac{n!}{2\cdot 5\cdot\ldots\cdot(3n-1)}=\sum_{n=1}^\infty \frac{n}{3^n}\int_0^1(1-x)^{n-1}x^{-1/3}\,dx=$$ $$=\sum_{n=1}^\infty \frac{n}{3^n}\int_0^1(1-u^{3/2})^{n-1}\frac32\,du=\frac12\int_0^1 \sum_{n=1}^\infty n\left(\frac{1-u^{3/2}}{3}\right)^{n-1}\,du.$$ Using the relation $$\sum_{n=1}^\infty nq^{n-1}=\frac{1}{(1-q)^2}$$ (which is just the derivative of $\sum\limits_{n=1}^\infty q^{n}=\frac{1}{1-q}$) with $q=\frac{1-u^{3/2}}{3}$ we get $$S=\frac12\int_0^1\frac{du}{(1-\frac{1-u^{3/2}}{3})^2}=\frac92\int_0^1\frac{du}{(2+u^{3/2})^2}=9\int_0^1 \frac{t\,dt}{(t^3+2)^2}.$$ Since we have antiderivative for this function of the form $$\int \frac{9t\,dt}{(t^3+2)^2}=\frac{3t^2}{2(t^3+2)}+\frac{1}{4\sqrt[3]{2}}\ln(t^3+2)-\frac{3}{4\sqrt[3]{2}}\ln(t+\sqrt[3]{2})+\frac{\sqrt3}{2\sqrt[3]{2}}\arctan\frac{\sqrt[3]{4}t-1}{\sqrt3}+C,$$ by the fundamental theorem of calculus $$S=\frac12+\frac{1}{4\sqrt[3]{2}}\ln3-\frac{3}{4\sqrt[3]{2}}\ln(1+\sqrt[3]{2})+\frac{\sqrt3}{2\sqrt[3]{2}}\arctan\frac{\sqrt[3]{4}-1}{\sqrt3}+\frac{\sqrt3}{2\sqrt[3]{2}}\arctan\frac{1}{\sqrt3}=$$ $$=\frac12+\frac{1}{4\sqrt[3]{2}}\ln(\sqrt[3]{2}-1)+\frac{\sqrt3}{2\sqrt[3]{2}}\left(\frac{\pi}{6}+\arctan\frac{\sqrt[3]{4}-1}{\sqrt3}\right).$$ Adding again omitted earlier first term $1$, we obtain an expression for required sum which is equivalent to one given by Mathematica.

Answer 2

Recall that the form of Gauss's hypergeometric series is given by $$_2F_1(a,b,c;x)=\sum_{k=0}^\infty \frac{(a)_k(b)_k}{(c)_k}\frac{x^k}{k!}$$ where $(n)_k=n(n+1)\cdots(n+k-1)$ is the rising factorial / Pochhammer symbol. Observe that the ratio of consecutive terms in this series is

$$\frac{(a)_{k+1}(b)_{k+1}/(c)_{k+1}\cdot x^{k+1}/(k+1)!}{(a)_{k}(b)_{k}/(c)_k\cdot x^k/k!}=\frac{(k+a)(k+b)}{(k+c)(k+1)}x.$$

By comparison, the ratio of consecutive terms in the series given by the OP is readily seen to be $$\dfrac{k+1}{3k+2}=\dfrac{(k+1)(k+1)}{(k+2/3)(k+1)}\cdot \dfrac{1}{3}.$$ Comparing these, we find that we need $(a,b,c,x)=(1,1,2/3,1/3)$. This can be verified by expanding the resulting hypergeometric function in WolframAlpha and taking $x=1/3$.

So all we need to do is compute is $_2F_1\left(1,1,\frac{2}{3};\frac13\right)$. Typically, evaluating a hypergeometric function can only be done numerically. In this case, however, WolframAlpha reports that hypergeometric series has a (complicated) closed form in terms of $x$ and so the result will be algebraic. As of yet I have not managed to verify for myself (by hand or by reference) the result given by the OP from WolframAlpha. But WolframAlpha does confirm that they are the same.