At the 1993 Tokyo University Exam, the following problem shocked high-schoolers. At the time of 1990’s, the Japanese curriculum used a very rough estimate of $\pi\simeq 3$ during calculations. However, the exam pretty much tested how the students did not merely memorize this approximation and truly understood the mathematics behind it.

Problem. Show that $\pi > 3.05$ .

Solution.

Since I do not know how the curriculum were at the time, I will provide my own approach which seems the simplest. We first note the geometric series

$$ \frac{1}{1+x^2}=1-x^2+x^4+... $$

for $\vert x \vert <1$. Now, integrating both sides leave

$$ \arctan(x)+C=x-\frac{x^3}{3}+\frac{x^5}{5}-... $$

with taking $x=0$ gives $C=0$ so that $\arctan(x)=x-x^3/3+x^5/5 ...$ , and letting $x=1$ gives

$$ , \pi=4(1-1/3+1/5-1/7+....)=\Sigma_{k=0}^{\infty}\frac{4\times(-1)^k}{2k+1} $$

, which is an alternating series with decaying magnitude $\vert a_n \vert>\vert a_{n+1}\vert$. Therefore, if we truncate the series at any even term, we observe $\pi>S_{even}$ where S denotes the partial sum up to the term of interest. So, we can just truncate at 1/17, leaving

$$ \pi>4(1-1/3+1/5-1/7-1/9-1/11+1/13-1/15+1/17))\simeq 3.25>3.05 $$

If anyone can share a more elementary and elegant solution let me know!