<s>[INST] 2 problems related to the number 2015 
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*Let $p=\underbrace{11\cdots1}_\text{2015}\underbrace{22\cdots2}_\text{2015}$. Find $n$, where $n(n+1) = p$

*Prove that $\frac{1}{2^2} + \frac{1}{3^2} + \cdots +  \frac{1}{2015^2} < \frac{2014}{2015}$
For 1, I tried dividing in various ways until I got a simpler expression, but no result. For 2, I know that $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$, but the proofs i found are over the elementary level the problem is aimed for. I also proved that it's smaller than $\frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{1024^2} < 1$, but that obviously is missing $\frac{1}{2015}$
 [/INST] $1)$ $n=\frac{10^{2015}-1}{3}=3...3$ with $2015$ $3's$ :
$$\frac{10^{2015}-1}{3}\times \frac{10^{2015}+2}{3}=\frac{10^{4030}+10^{2015}-2}{9}=10^{2015}\times \frac{10^{2015}-1}{9}+2\times \frac{10^{2015}-1}{9}=10^{2015}R_{2015}+2\times R_{2015}$$
$2)$ follows from $\frac{1}{2^2}+...+\frac{1}{2015^2}<\sum_{j=2}^{\infty} \frac{1}{j^2}=\frac{\pi^2}{6}-1<\frac{2014}{2015}$
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