3. Let $x=\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\cdots+\sqrt{1+\frac{1}{2012^{2}}+\frac{1}{2013^{2}}}$. Find the value of $x-[x]$, where $[x]$ denotes the greatest integer not exceeding $x$.
(1 mark)
設 $x=\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\cdots+\sqrt{1+\frac{1}{2012^{2}}+\frac{1}{2013^{2}}}$ 。求 $x-[x]$ 的值 （這裡 $[x]$ 表示不超過 $x$ 的最大整數）。
3. $\frac{2012}{2013}$
3. Computing the first few terms gives $\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}=\frac{3}{2}=1+\frac{1}{1 \times 2}, \sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}=\frac{7}{6}=1+\frac{1}{2 \times 3}$, $\sqrt{1+\frac{1}{3^{2}}+\frac{1}{4^{2}}}=\frac{13}{12}=1+\frac{1}{3 \times 4}$. It is thus reasonable to guess that
$$
\sqrt{1+\frac{1}{k^{2}}+\frac{1}{(k+1)^{2}}}=1+\frac{1}{k(k+1)}
$$
and this indeed is true (and can be verified algebraically). Thus we need only consider the fractional part of
$$
\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{2012 \times 2013} .
$$

Since $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$, the above sum becomes
$$
\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{2012}-\frac{1}{2013}\right)=\frac{1}{1}-\frac{1}{2013}=\frac{2012}{2013}
$$
which is the answer to the question.