Sum from n=1 to ∞ of (-1)^n(sin(7π/n)) is the series convergent or divergent?

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the 7 is throwing me off I'm unsure of how to go about it

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Answer 1 · 2018-03-01T09:20:10

The series:

#sum_(n=1)^oo (-1)^nsin((7pi)/n)#

is convergent.

This is an alternating series in the form:

#sum_(n=1)^oo (-1)^n a_n#

where #a_n = sin((7pi)/n)#.

First we note that:

#(1) " " lim_(n->oo) a_n = 0#

Then consider that for #n>14# we have:

#0 < (7pi)/n < pi/2#

so that:

#a_n > 0#

and, as in the interval #(0,pi/2)# the function #sinx# is strictly increasing:

#(7pi)/(n+1) < (7pi)/n => sin((7pi)/(n+1)) < sin((7pi)/n)#

that is:

#(2) " " a_(n+1) < a_n#

Based on Leibniz's theorem then the conditions #(1)# and #(2)# are sufficient to prove that the series is convergent.