Mind the 3 decimal place

$\begin{aligned} & \frac{1}{1}+\frac{1}{4}+\frac{1}{9} + \frac{1}{16}+ \frac{1}{25}+\frac{1}{36}\cdots > \frac{1}{1}+\frac{1}{2}+\frac{1}{8}+\frac{1}{48}+\frac{1}{384} + \frac{1}{3840}\cdots \\& \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2} + \frac{1}{4^2}+ \frac{1}{5^2}+\frac{1}{6^2}\cdots > \frac{1}{1}+\frac{1}{2.1}+\frac{1}{4.2}+\frac{1}{8.6}+\frac{1}{24.16} + \frac{1}{64.120}\cdots \\& \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2} + \frac{1}{4^2}+ \frac{1}{5^2}+\frac{1}{6^2}\cdots > \frac{1}{1}+\frac{1}{2.1!}+\frac{1}{2^2.2!}+\frac{1}{2^3.3!}+\frac{1}{2^4.4!} + \frac{1}{2^5.5! }\cdots\end{aligned}$ $\begin{aligned} &\displaystyle \sum_{n= 1}^{\infty} \frac{1}{n^2} >\displaystyle \sum_{k=0}^{\infty}\frac{1}{2^k k!} \\& \frac{\pi^2}{6} > e^{\frac{1}{2}} \Rightarrow \pi^4 > 36e \end{aligned}$ Hence the answer is $\text{False}$ .