Let f(x)=4π⋅(sinx+13sin(3x)+15sin(5x)+…). If for x=π2, we have
f(x)=4π(1−13+15−17+…)=1
then obviously :
1−13+15−17+⋯=π4
Now how can we prove that:
π28=1+132+152+172+…
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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