Limit of $a_n = \frac{1}{1^3\cdot 1}+\frac{1}{1^3\cdot 2+2^3\cdot 1}+\cdots+\frac{1}{1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1}$ Let 
$$a_n = \frac{1}{1^3\cdot 1}+\frac{1}{1^3\cdot 2+2^3\cdot 1} + \cdots +\frac{1}{1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1}$$
Does this sequence converge to a simple number?
My thought was to compute each denominator:
$$1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1=(n+1)\sum_{k=1}^n k^3-\sum_{k=1}^nk^4$$
and this are known, I find $\frac{1}{60}n(n+1)(2n+1)(3n^2+6n+1)$. But can we find maybe a closed formula for $a_n$ after this?

Continuing in a different way, you can express the limit in terms of the digamma function as follows:
$$
\mathop {\lim }\limits_{n \to  + \infty } 30\left[ {\sum\limits_{k = 1}^n {\frac{1}{k}}  + \sum\limits_{k = 1}^n {\frac{1}{{k + 1}}}  + \sum\limits_{k = 1}^n {\frac{1}{{k + 2}}}  - \frac{3}{2}\sum\limits_{k = 1}^n {\frac{1}{{k + 1 + \sqrt {2/3} }}}  - \frac{3}{2}\sum\limits_{k = 1}^n {\frac{1}{{k + 1 - \sqrt {2/3} }}} } \right]
\\
 = \mathop {\lim }\limits_{n \to  + \infty } 30\left[ {\frac{1}{2} - \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \frac{3}{2}\sum\limits_{k = 1}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 + \sqrt {2/3} }}} \right]}  + \frac{3}{2}\sum\limits_{k = 1}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 - \sqrt {2/3} }}} \right]} } \right]
\\
 = \mathop {\lim }\limits_{n \to  + \infty } 30\left[ {6 + \frac{1}{2} - \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \frac{3}{2}\sum\limits_{k = 0}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 + \sqrt {2/3} }}} \right]}  + \frac{3}{2}\sum\limits_{k = 0}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 - \sqrt {2/3} }}} \right]} } \right]
\\
 = 30\left[ {6 + \frac{1}{2} + \frac{3}{2}\left( {\psi \left( {\sqrt {2/3} } \right) + \gamma } \right) + \frac{3}{2}\left( {\psi \left( { - \sqrt {2/3} } \right) + \gamma } \right)} \right]
\\
 = 195 + 90\gamma  + 45\left( {\psi \left( {\sqrt {2/3} } \right) + \psi \left( { - \sqrt {2/3} } \right)} \right)
\\
 = 195 + 90\gamma  + 45\pi \cot (\pi \sqrt {2/3} ) + 45\sqrt {\frac{3}{2}}  + 90\psi (\sqrt {2/3} ).
$$
