Test for convergence

Question

anonymous · Answer

$$\sum_{n=1}^{\infty}\sin^3\frac{1}{n}$$

shubhamsrg · Answer

well since at n=infintiy , we get 0 , the series converges for sure,,but am not too sure about the ans..

anonymous · Answer

shubhamsrg how can you be so sure ?

shubhamsrg · Answer

since height at infinity is 0,,wont that be satisfactory ?

anonymous · Answer

yes, we can say that the series conv. by intuition, but the problem is we need to prove that using convergence tests

anonymous · Answer

if im not wrong limit 0 at infinity is not enough for it..

shubhamsrg · Answer

hmmm..

anonymous · Answer

wolframalpha says that comparison test works, but I have no idea what should I compare

anonymous · Answer

look at 1/n .. not converge

anonymous · Answer

decreasing is not enough as well

shubhamsrg · Answer

will this help :

we know sinx <= x for x>0 ..

thus
sin (1/n^2) <= 1/n^2 

since the RHS is converging,surely the LHS must be converging..

will that be right?

anonymous · Answer

i think it is good .. i wanted to use the comparison test by my self but couldnt find how

calculusfunctions · Answer

I will be silent for now but will say that only one person has been on the right track thus far, but has failed to give a reason or a theorem to back it up. Any guesses who the medal might go to?

anonymous · Answer

I think |dw:1350639241956:dw| is finite.