Summation of a series with certain property The sequence $\{a_n\}$ has the property that $a_1 + a_2+\cdots+a_n = n ^3~~\forall n$. Compute the value of $\frac{1}{a_2} + \frac{1}{a_3}+\ldots+ \frac{1}{a_{2014-1}}$. 
I know that I have to somehow a arrange the denominators in such a order that I could take advantage of the property. To add them, I have to multiply the denominators to obtain a common denominators.
 A: First observe that
$$
a_n=(a_1+\cdots+a_{n-1}+a_n)-(a_1+\cdots+a_{n-1})=n^3-(n-1)^3=3n^2-3n+1.
$$
Thus
\begin{align}
\frac{1}{a_2-1}+\cdots+\frac{1}{a_{2014}-1}&=\frac{1}{3 (2^2-2)}+\cdots+\frac{1}{3 (2014^2-2014)}=\frac{1}{3}\sum_{k=2}^{2014}\frac{1}{k(k-1)}\\ &=\frac{1}{3}
\sum_{k=2}^{2014}\left(\frac{1}{k-1}-\frac{1}{k}\right)
=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{2014}\right)=\frac{671}{2014},
\end{align}
since
$$
\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}.
$$
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left( #1 \right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large a}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$$
a_{n} = \pars{a_{1} + \cdots + a_{n}} - \pars{a_{1} + \cdots + a_{n - 1}}
=n^{3} - \pars{n - 1}^{3} = 3n^{2} - 3n + 1
$$

$$
\sum_{k = 1}^{m}{1 \over a_{k}}
=\sum_{k = 1}^{m}{1 \over 3k^{2} - 3k + 1}
={1 \over 3}\sum_{k = 1}^{m}{1 \over \pars{k - z}\pars{k - z*}}\,,\qquad
z \equiv {1 \over 6}\pars{3 + \root{3}\ic}
$$

\begin{align}
\sum_{k = 1}^{m}{1 \over a_{k}}
&={1 \over 3}\,{1 \over z - z^{*}}
\sum_{k = 1}^{m}\pars{{1 \over k - z} - {1 \over k - z^{*}}}
={1 \over 3}\,{1 \over 2\ic\,\Im\pars{z}}
2\ic\,\Im\sum_{k = 1}^{m}{1 \over k - z}
\\[3mm]&={1 \over 3}\,{1 \over \root{3}/6}\Im\sum_{k = 1}^{m}{1 \over k - z}
={2\root{3} \over 3}\,\Im\sum_{k = 1}^{m}{1 \over k - z}
={2\root{3} \over 3}\,\Im\bracks{\Psi\pars{1 + m - z} - \Psi\pars{1 - z}}
\end{align}

$$\color{#00f}{\large%
\sum_{k = 2}^{m}{1 \over a_{k}}
=
{2\root{3} \over 3}\,\Im\bracks{%
\Psi\pars{m + \half - {\root{3} \over 6}\,i} - \Psi\pars{\half - {\root{3}
\over 6}\,i}} - 1}
$$

$\ds{\Psi\pars{z}}$ is the Digamma Function.
