Find the integer closest to $ a - b$

Question

Let $$a = \frac{1^{2}}{1} + \frac{2^{2}}{3} + \frac{3^{2}}{5} + \ldots + \frac{1001^{2}}{2001}.$$

Let $$b = \frac{1^{2}}{3} + \frac{2^{2}}{5} + \frac{3^{2}}{7} + \ldots + \frac{1001^{2}}{2003}.$$

Then, what is the closest integer of $(a - b)$?

Answer 1

We have $$\begin{align}a-b&=\sum_{k=1}^{1001}\frac{k^2}{2k-1} -\sum_{k=1}^{1001}\frac{k^2}{2k+1}\\ &=\frac{1^2}1-\frac{1001^2}{2003}+\sum_{k=1}^{1000}\frac{(k+1)^2-k^2}{2k+1}\\ &=\frac{1^2}1-\frac{1001^2}{2003}+\sum_{k=1}^{1000}1\\ &=1001-\frac{1001}{2+\frac1{1001}}\\ &\approx 1001-\frac{1001}2\cdot\left(1-\frac1{2002}\right)\\ &=500\tfrac34\end{align}$$-

Answer 2

\begin{align} & a-b=\sum\limits_{i=1}^{1001}{\left( \frac{{{i}^{2}}}{2i-1}-\frac{{{i}^{2}}}{2i+1} \right)=}\sum\limits_{i=1}^{1001}{\frac{2{{i}^{2}}}{4{{i}^{2}}-1}=}\frac{1}{2}\sum\limits_{i=1}^{1001}{\frac{4{{i}^{2}}-1}{4{{i}^{2}}-1}+\frac{1}{2}\sum\limits_{i=1}^{1001}{\frac{1}{4{{i}^{2}}-1}}} \\ & a-b=\frac{1001}{2}+\frac{1}{4}\sum\limits_{i=1}^{1001}{\left( \frac{1}{2i-1}-\frac{1}{2i+1} \right)}=\frac{1001}{2}+\frac{1}{4}\left( 1-\frac{1}{2003} \right) \\ \end{align}