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

Question

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

Isaiah Allen · Accepted Answer

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

#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# The sequence is convergent as a result. 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.

Elijah Abram · Answer

undefined The interval of convergence for the series $ \sum_{n=1}^{\infty} \frac{\sin(nx)}{n} $ is $[-1, 1]$.