143. a) Derive from the identities of problems 140a) and b) the Euler's formula

$$
\frac{\pi^{2}}{6}=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots
$$

b) What is the sum of the infinite series

$$
1+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\frac{1}{4^{4}}+\ldots ?
$$
143. a) Since for angles of the first quadrant $\operatorname{cosec} \alpha > \frac{1}{\alpha} > \operatorname{ctg} \alpha$ (see problem 136a)), it follows from the identities of problems 140a) and b) that

$$
\begin{aligned}
& \frac{m(2 m-1)}{3}< \\
& <\left(\frac{2 m+1}{\pi}\right)^{2}+\left(\frac{2 m+1}{2 \pi}\right)^{2}+\ldots+\left(\frac{2 m+1}{m \pi}\right)^{2}<\frac{m(2 m+2)}{3}
\end{aligned}
$$

$$
\begin{array}{r}
\frac{\pi^{2}}{6}\left(1-\frac{1}{2 m+1}\right)\left(1-\frac{2}{2 m+1}\right)<1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{m^{2}}< \\
<\frac{\pi^{2}}{6}\left(1-\frac{1}{2 m+1}\right)\left(1+\frac{1}{2 m+1}\right)
\end{array}
$$

But since the extreme terms of the last double inequality tend to $\frac{\pi^{2}}{6}$ as $m \rightarrow \infty$, we have

$$
\lim _{m \rightarrow \infty}\left\{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{m^{2}}\right\}=\frac{\pi^{2}}{6}
$$

b) Let's find the sum of the squares of the roots of the equation from problem 139a). By equating the coefficients of $x^{m-2}$ on the left and right sides in the identity $\left({ }^{*}\right)$ from the solution of problem 140a), we find that the sum of all possible pairwise products of the roots of the equation is

$$
\frac{C_{2 m+1}^{5}}{A}=\frac{C_{2 m+1}^{5}}{C_{2 m+1}^{1}}=\frac{m(2 m-1)(m-1)(2 m-3)}{30}
$$

Since the sum of the squares of any quantities is equal to the square of their sum, decreased by twice the sum of all possible pairwise products of these quantities, we get:

$$
\begin{aligned}
\operatorname{ctg}^{4} \frac{\pi}{2 m+1}+\operatorname{ctg}^{4} & \frac{2 \pi}{2 m+1}+\operatorname{ctg}^{4} \frac{3 \pi}{2 m+1}+\ldots+\operatorname{ctg}^{4} \frac{m \pi}{2 m+1}= \\
& =\frac{m^{2}(2 m-1)^{2}}{9}-\frac{m(2 m-1)(m-1)(2 m-3)}{15}= \\
& =\frac{m(2 m-1)\left(4 m^{2}+10 m-7\right)}{45}
\end{aligned}
$$

Further, since $\operatorname{cosec}^{4} \alpha=\left(\operatorname{ctg}^{2} \alpha+1\right)^{2}=\operatorname{ctg}^{4} \alpha+2 \operatorname{ctg}^{2} \alpha+1$, we have:

$$
\begin{aligned}
& \operatorname{cosec}^{4} \frac{\pi}{2 m+1}+\operatorname{cosec}^{4} \frac{2 \pi}{2 m+1}+\ldots+\operatorname{cosec}^{4} \frac{m \pi}{2 m+1}= \\
& \quad=\frac{m(2 m-1)\left(4 m^{2}+10 m-7\right)}{45}+\frac{2 m(2 m-1)}{3}+m= \\
& \quad=\frac{8 m(m+1)\left(m^{2}+m+3\right)}{45}
\end{aligned}
$$

Now, similarly to the solution of problem a), we obtain the following double inequality:

$$
\begin{aligned}
& \frac{m(2 m-1)\left(4 m^{2}+10 m-7\right)}{45}< \\
& <\left(\frac{2 m+1}{\pi}\right)^{4}+\left(\frac{2 m+1}{2 \pi}\right)^{4}+\left(\frac{2 m+1}{3 \pi}\right)^{4}+\ldots+\left(\frac{2 m+1}{m \pi}\right)^{4}< \\
& <\frac{8 m(m+1)\left(m^{2}+m+3\right)}{45}
\end{aligned}
$$

and

$$
\begin{aligned}
& \frac{\pi^{4}}{90}\left(1-\frac{1}{2 m+1}\right)\left(1-\frac{2}{2 m+1}\right)\left(1+\frac{3}{2 m+1}-\frac{13}{(2 m+1)^{2}}\right)< \\
& <1+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots+\frac{1}{m^{4}}<\frac{\pi^{4}}{90}\left(1-\frac{1}{(2 m+1)^{2}}\right)\left(1+\frac{11}{(2 m+1)^{2}}\right)
\end{aligned}
$$

from which, passing to the limit as $m \rightarrow \infty$, we get:

$$
1+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\frac{1}{4^{4}}+\ldots=\frac{\pi^{4}}{90}
$$

Remark. By similarly determining the sum of cubes, fourth powers, and so on of the roots of the equation from problem 139a), we can obtain the following series of formulas:

$$
\begin{gathered}
1+\frac{1}{2^{8}}+\frac{1}{3^{8}}+\frac{1}{4^{8}}+\ldots=\frac{\pi^{8}}{945} \\
1+\frac{1}{2^{8}}+\frac{1}{3^{8}}+\frac{1}{4^{8}}+\ldots=\frac{\pi^{8}}{9450} \\
1+\frac{1}{2^{10}}+\frac{1}{3^{10}}+\frac{1}{4^{10}}+\ldots=\frac{\pi^{10}}{93555} \\
1+\frac{1}{2^{12}}+\frac{1}{3^{12}}+\frac{1}{4^{12}}+\ldots=\frac{691 \pi^{12}}{638512875} \\
\ldots \ldots \ldots . \ldots . . . . . .
\end{gathered}
$$

All these formulas were first obtained by L. Euler.