The forum discussion centers on determining the convergence or divergence of the series Σ (n sin n) / (n^3 + 1) from n = 1 to infinity. Participants suggest using the comparison test, particularly comparing with (n sin n) / n^3, and discuss the behavior of the numerator and denominator as n approaches infinity. The conclusion drawn is that the series converges to 0, as the simplified form sin n / n^2 approaches 0, despite the oscillatory nature of sin n.
PREREQUISITESMathematics students, educators, and anyone interested in series convergence, particularly those studying calculus or advanced mathematical analysis.
-EquinoX- said:Homework Statement
the sum of (n sin n)/(n^3+1)
n from 1 to infinity
Homework Equations
The Attempt at a Solution
what I have done so far is using the comparison method by comparing this with n sin n/ n^3, but I don't know what's the next step and even if my comparison is correct or not
-EquinoX- said:that's where I was stuck. I think it doesn't go anywhere therefore it doesn't exist. Am I right??
-EquinoX- said:well because sin n varies between -1 and 1. This what makes me think that the limit does not converges to some value
-EquinoX- said:it goes to infinity, so therefore it goes to 0?? Am I right?
-EquinoX- said:yeah If I replace it with either -1 or 1 both will most likely to reach 0 from both sides, from the negative sides and the positive sides.
-EquinoX- said:well we can simplify [n * sin n]/n^3 into sin n/n^2 right?? therefore the answer is 0?
-EquinoX- said:what do you mean by deduce?
HallsofIvy said:Excuse me, but the original question asked for the SUM of \frac{nsin(n)}{n^3+1}, not the limit of the sequence.