Friday, 11 December 2015

calculus - Show by comparison that sumlimitsin=1nftysin(frac1n) diverges?




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 limnanbn 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=1nlimt0sintt=1



The result now follows by the limit comparison test. :)


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