We know that, if $b_k$ is monotone decreasing and $\lim b_k =0$, then $\sum b_k \sin kt $ is convergent for all $t\in R $.
IF we change the condition from $b_1 \geq b_2 \geq ....\geq 0$
to
$b_N+1 \geq b_N+2 \geq ....\geq 0$, i.e, $b_k$ is monotone decreasing after some $N$.
Does this test still hold ?
I my intuition is that it still holds.
Any hint?
Hint $$\sum_{k=0}^\infty b_k \sin kt = \sum_{k=0}^N b_k \sin kt +\sum_{k=N+1}^\infty b_k \sin kt = \sum_{k=0}^N b_k \sin kt + \sum_{k=0}^\infty a_k \sin ((N+1+k)kt) $$ The first sum is a finit number and $a_k=b_{N+1+k}$
then you can use $\sin(a+b)= \sin a \cos b+ \sin b \cos a$ and the test shown below