By using the integral test, I know that ∞∑n=1sin(1n)
diverges. However, how would I show that the series diverges using the limit comparison test? Would I simply let ∞∑n=1an=∞∑n=1bn=∞∑n=1sin(1n) and then take limn→∞anbn to show the series diverges (assuming the limit converges to some nonnegative, finite value)?
Answer
Notice xn=1n, then ∑1n, the harmonic series, we all know is divergent. Now,
limsin(1n)1n=t=1nlimt→0sintt=1
The result now follows by the limit comparison test. :)
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