\input amstex
\input cyracc.def
\font\tencyr=wncyr10
\def\cyr#1{{\tencyr\cyracc#1}}

\cyr{Parsevalovom jednakosh\'cu se dobija:}
$$\sum\limits_{k=0}^\infty\frac{1}{(2k+1)^2}=\frac{\pi^2}{8}.$$

\cyr{Takodje sam dobio:}
$$\sum\limits_{k=0}^\infty\frac{(-1)^k}{2k+1}=\frac{\pi}{4}.$$

\cyr{I sad:}
$$\sum\limits_{k=0}^\infty\frac{(-1)^k}{4k+1}=?$$

\bye