does this series converge ?

Question

Summation from j=1 to infinity:

does sin(1/j) converge?
how about sin(1/j)^2 ?

any help would be appreciated

Answer 1

I don't think the above answerers were paying too much attention. The sequence {sin (1/j)} does in fact converge to zero, but the series [j=1, ∞]∑sin (1/j) diverges. The easiest way to see this is to note that sin (1/j) > 1/(2j) for all j≥1, and [j=1, ∞]∑1/(2j) diverges (it is a harmonic series).

[j=1, ∞]∑sin² (1/j) converges, since sin (1/j) < 1/j for all j≥1, thus sin² (1/j) < 1/j², so this series is bounded above by [j=1, ∞]∑1/j² = π²/6

Answer 2

Yes because 1/j converges to 0 as j converges to infinity.
And sinx is continuous fanction.
So sin(1/j) converges to sin0=0 as j converges to infinity.

Answer 3

I think so for both