Closed form for $\sum_{k=1}^{n} \frac{k}{1.4.7.10\ldots(3k-2)}$

Question

I need to find Closed form for $$\sum_{k=1}^{n} \dfrac{k}{1.4.7.10.....(3k-2)}$$

Let $S_n=\displaystyle\sum_{k=1}^n \dfrac{k}{1\cdot4\cdot7\cdot10\cdots(3k-2)}$.

Then, $S_1=1$, $S_2=\dfrac{6}{1\cdot4}$, $S_3=\dfrac{45}{1\cdot4\cdot7}$. Are we getting at something ?

By ratio test, it can be seen that the series is convergent but is it possible to find an expression for the $n$the term of te sequence of $n$th partial sums ?

Answer 1

Suppose,

$$f(x)=\sum_{k=0}^{n} \frac{x^{(3k-2)}}{1.4.7...(3k-2)}$$

So, $$f'(x)=1+x^2\sum_{k=0}^{n-1} \frac{x^{(3k-2)}}{1.4.7...(3k-2)} -a_nx^{3n}=1+f(x)-a_nx^{(3n-2)}$$

(Where $a_n=\frac{1}{1.4.7....(3n-2)}$)

So we got the differential equation,

$\frac{\text{d}y(x)}{\text{d}x}-x^2y=1-a_nx^{3n}$ with the boundary condition $y(0)=0$.

Now, it is easy to see the original sum is $S_n=\frac{\text{d}}{\text{d}x^3}x^2y(x)|_{x=1}=\frac{1}{3}(\frac{\text{d}}{\text{d}x}x^2y(x))|_{x=1}$

For, large $n$, say $n≈20$, $S_n≈S_{\infty}$.

So, the differential equation will be replaced by a simpler version,

$\frac{\text{d}y(x)}{\text{d}x}-x^2y=1$ with boundary condition $y(0)=0$.

So, $y(x)=e^{\frac{x^3}{3}}\int_{0}^{x} e^{-\frac{t^3}{3}} \text{d}t$

So, $\frac{1}{3}(\frac{\text{d}}{\text{d}x}x^2y(x))|_{x=1}$

$=\frac{1}{3}((\frac{\text{d}}{\text{dx}}(x^2e^{-\frac{x^3}{3}})\int_{0}^{x} e^{-\frac{t^3}{3}} dt)|{x=1} + (x^2)|{x=1})$

$=\frac{1}{3}(3e^{\frac{1}{3}}\int_{0}^{1} e^{-\frac{t^3}{3}} dt+ 1)$

$≈\frac{1}{3}(3e^{\frac{1}{3}}(0.924023)+1) ≈1.62291131$.

Answer 2

You want to compute $$S_n=\sum_{k=1}^n \frac{k}{\prod_{i=1}^k (3i-2)}$$ $$\frac{k}{\prod_{i=1}^k (2i-1)}=\Gamma \left(\frac{1}{3}\right)\,\frac{ k\, }{3^k\,\Gamma \left(k+\frac{1}{3}\right)}$$ A CAS gives a long closed form expression (I shall not reproduce it here; it contains $14$ gamma functions !). Fortunately, this simplifies and can write $$S_n=\frac{1}{3} \left(1-\sqrt[3]{e} E_{\frac{2}{3}}\left(\frac{1}{3}\right)\right)+\Gamma \left(\frac{1}{3}\right)\frac{ \sqrt[3]{e} E_{-(n+\frac{1}{3})}\left(\frac{1}{3}\right)-3 n-4}{3^{n+2}\Gamma \left(n+\frac{4}{3}\right)}$$ where appear exponential integral functions.

This generates the sequence $$\left\{1,\frac{3}{2},\frac{45}{28},\frac{227}{140},\frac{5907}{3640},\frac{47259}{29120 },\frac{1795849}{1106560},\frac{2822049}{1738880},\frac{6907113}{4256000},\frac{1382 8040231}{8520512000},\frac{857338494333}{528271744000},\frac{14574754403667}{8980619 648000},\frac{1078531825871371}{664565853952000},\frac{21570636517427427}{1329131707 9040000},\frac{7162450735516443}{4413333084160000}\right\}$$ which is not recognized by $OEIS$.