# Computing the value of a fraction involving summation

can we compute this term without using calculator?

$$\frac{1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{997^2}+\frac{1}{999^2}-\frac{1}{1002^2}-\frac{1}{1004^2}-\frac{1}{1006^2}-...-\frac{1}{1998^2}-\frac{1}{2000^2}}{1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}}$$

I can't see any pattern.

I tried to add and substract as mentioned in the comments but can't see how it helps.

thanks.

• In the numerator add and substrate the even terms present in the denominator (1/2^2 +1/4^2+... ). – incognito Oct 27 '15 at 15:09
• @incognito i tried that now but can't see how it helps. – tammy Oct 27 '15 at 16:28
• @tammy: Now factor $4$ in the denominator of each. – Lucian Oct 28 '15 at 1:43

The given fraction $$\frac{1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{997^2}+\frac{1}{999^2}-\frac{1}{1002^2}-\frac{1}{1004^2}-\frac{1}{1006^2}-...-\frac{1}{1998^2}-\frac{1}{2000^2}}{1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}}$$ can be rewritten as $$\frac{\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}\right)-\left(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{1000^2}+\frac{1}{1002^2}+...+\frac{1}{1998^2}+\frac{1}{2000^2}\right)}{1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}}$$ by adding and subtracting $\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{1000^2}$ to the numerator of the fraction. The above fraction can now be written as: $$1 - \frac{\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{1000^2}+\frac{1}{1002^2}+\frac{1}{1004^2}+...+\frac{1}{1998^2}+\frac{1}{2000^2}}{1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}}$$
$$= 1 - \frac{\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}\right)}{1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}} = 1 - \frac{1}{2^2} = \frac{3}{4}.$$