What is the interval of convergence of #sum_1^oo sin(nx)/n #?

1 Answer
Apr 6, 2017

#sum_(n=1)^(oo)sin(nx)/n# is conditionally convergent by the alternating series test and the harmonic series.

Explanation:

#sum_(n=1)^oo frac{sin(nx)}{n}#

We can think of this as if it were an alternating series , because the numerator, #sin(nx)#, will only oscillate between #-1# and #1#.

The alternating series test states that we need to prove that:
#color(blue)("I " lim_(n->oo)a_n =0 #
#color(blue)(" where "a_n" excludes the alternating part of the series."#

#color(blue)("II "a_n" is monotonically decreasing, " a_(n+1)<=a_n )#

#"Let " a_n=1/n#.

#color(red)("I. ")lim_(n->oo)(1/n)=0#

#color(red)("II. ") 1/(n+1)<=1/n#

Therefore, the sequence is convergent.

To determine whether it has absolute or conditional convergence, test the convergence of:
#sum_(n=1)^(oo)1/n#
We know this is divergent by the harmonic series.

Therefore, #sum_(n=1)^(oo)sin(nx)/n# is conditionally convergent by the alternating series test and the harmonic series.