# Science:Math Exam Resources/Courses/MATH101/April 2015/Question 05 (b)/Solution 1

Since ${\displaystyle a_{n}={\frac {1}{(2n+1)^{2}}}}$ is positive for any ${\displaystyle n\geq 1}$, monotonically decreasing, and converges to 0, applying the Alternating Series Estimation Theorem, we have ${\displaystyle |S-S_{n}|\leq a_{n+1}={\frac {1}{(2(n+1)+1)^{2}}}={\frac {1}{(2n+3)^{2}}}.}$
Now, it is enough to find the smallest ${\displaystyle n}$ such that ${\displaystyle {\frac {1}{(2n+3)^{2}}}\leq {\frac {1}{100}}.}$
By taking square roots on both side, the inequality is equivalent with ${\displaystyle {\frac {1}{2n+3}}\leq {\frac {1}{10}}}$ and hence with ${\displaystyle 10\leq 2n+3}$. Therefore, we can easily obtain the smallest natural number ${\displaystyle n}$ as 4 satisfying ${\displaystyle 10\leq 2n+3}$.
The answer: ${\displaystyle n=\color {blue}4}$ and the partial sum is ${\displaystyle S_{4}=\color {blue}{\frac {1}{3^{2}}}-{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}-{\frac {1}{9^{2}}}}$