Problem - 3172
Let $S_n$ be the sum of first $n$ terms in sequence $\{a_n\}$ where $$a_n=\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}$$ Find $\lfloor{S_n}\rfloor$ where the floor function $\lfloor{x}\rfloor$ returns the largest integer not exceeding $x$.
Firstly, we note that $$1+\frac{1}{n^2}+\frac{1}{(n+1)^2}=\Big(1+\frac{1}{n}-\frac{1}{n+1}\Big)^2$$
This identity holds because $$\Big(1+\frac{1}{n}-\frac{1}{n+1}\Big)^2 = 1+\frac{1}{n^2}+\frac{1}{(n+1)^2}+2\times\Big(\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n(n+1)}\Big)$$
and$$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$
Therefore
$$a_n=1+\frac{1}{n}-\frac{1}{n+1}$$
$$S_n=\Big(1+1-\frac{1}{2}\Big)+\Big(1+\frac{1}{2}-\frac{1}{3}\Big)+\cdots+\Big(1+\frac{1}{n}-\frac{1}{n+1}\Big)=n+\Big(1-\frac{1}{n+1}\Big)$$
$$\therefore 0 < 1-\frac{1}{n+1}<1 \implies \lfloor{S_n}\rfloor = \boxed{n}$$