<s>[INST] 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.
 [/INST] 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
$$
\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}.
$$
</s>