2020 EJC P1 Q10

The diagram shows the graph of \(y=\frac{1}{{{x}^{2}}+1}\) when \(x>0\).

1. Evaluate \(\displaystyle\displaystyle \displaystyle\int_{k}^{k+1}{\frac{1}{{{x}^{2}}+1}\mathrm{d}x}\) for \(k>0\), leaving your answer in terms of \(k\).[2]
2. By considering appropriate rectangles on the interval \(\left[ k,k+1 \right]\) for the curve \(y=\frac{1}{{{x}^{2}}+1}\), show that
   \(\frac{1}{{{\left( k+1 \right)}^{2}}+1}<{{\tan }^{-1}}\left( k+1 \right)-{{\tan }^{-1}}k<\frac{1}{{{k}^{2}}+1}\) for \(k\in {{\mathbb{Z}}^{+}}\).[2]
3. Use the identity \(\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}\) to show that
   \({{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\frac{x-y}{1+xy}\), where \(x>y>0\).[2]
4. By considering parts (ii) and (iii), prove by the method of differences that
   \(\sum\limits_{k=1}^{n}{\frac{1}{{{\left( k+1 \right)}^{2}}+1}}<{{\tan }^{-1}}\left( \frac{n}{n+2} \right)<\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+1}}\)[4]