MHB What is the value of $a_{2015}$ in a sequence with given conditions?

[Ackbach]

Here is this week's POTW:

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Let $a_1, \dots, a_{2015}$ be real numbers such that $a_1=2015$ and
$$\sum_{j=1}^{n}a_{j}=n^2 \, a_n$$
for all $1\le n \le 2015$. Compute $a_{2015}$.

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[Ackbach]

Congratulations to kiwi, lfdahl, Opalg, kaliprasad, and MarkFL for their correct solutions! Quite the embarrassment of riches this week. As all the solutions are quite good, it was difficult to pick one. So, I employed a random number generator to get a number between 1 and 5, and the result was 2. So, I include lfdahl's solution below:

Spoiler

The sum $\sum_{j=1}^{n}a_j$ can be expressed in terms of $a_1$:
\[n = 1: \;\; a_1 = 2015\\\\
n = 2: \;\; a_1+a_2 = \frac{2^2}{2^2-1}a_1\\\\
n = 3: \;\; a_1+a_2+a_3 = \frac{3^2}{(3^2-1)}\cdot \frac{2^2}{(2^2-1)}\cdot a_1, \: \: \: ... etc.\]
In general: \[ \sum_{j=1}^{n}a_j = \prod_{k=2}^{n}\frac{k^2}{(k^2-1)}\cdot a_1=\gamma_n\cdot a_1 \]

$\gamma_n$ is a telescoping product:

\[\gamma_n = \prod_{k=2}^{n}\frac{k^2}{(k-1)(k+1)}=\frac{2}{2-1}\cdot \frac{n}{n+1}=\frac{2n}{n+1}\]

The term, $a_{2015}$, I am looking for, can now be expressed as follows:

\[ a_{2015}=\frac{1}{2015^2-1}\cdot \sum_{j=1}^{2014}a_j=\frac{1}{2015^2-1}\cdot \gamma_{2014}\cdot a_1\]

\[ =\frac{1}{2015^2-1}\cdot \frac{2\cdot 2014}{2015}\cdot 2015\]

\[=\frac{2\cdot 2014}{2014\cdot 2016} = \frac{1}{1008}\]