let:
anā=nĪ£(1/n)āā£sin(nr)ā£
bnā=nĪ£(1/n)ā
ā£sin(nr)ā£ā¤1 , then:
anāā¤bnā
also:
bnāā¤ā(1/n2)
So, since ā(1/n2) converges, bnā=nĪ£(1/n)ā converges as well
so, ābnā converges
then, āanā converges
so, series ā(1+1/2+...+1/n)nsin(nr)ā converges