Convergence of sin(log x)

Question

I have to show that the series $ {x_n} = { \sin (\log \ n ) }$ diverges.

If I start of with $ x_n - x_{n-1} $ that doesn't get me anywhere.

It is fairly obvious that $ \log n $ diverges. Can I use that and the fact that $ \sin x$ does not converge to say that the composition of these two series diverges as well?

Do you mean the series $\lim_{n \to \infty} \sum_{i = 1}^n x_i$ or the sequence $(x_n)_n$? — P. Senden, Apr 15 '18 at 08:39

score 3 · Answer 1 · answered Apr 15 '18 at 08:38

3

In general, you cannot use that if a sequence $(a_n)_n$ diverges, the sequence $\sin(a_n)_n$ also diverges. Consider for example $a_n = 2πn$. Then $(a_n)_n$ is clearly divergent, but $\sin(a_n)_n$ is the zero sequence, which trivially converges.

answered Apr 15 '18 at 08:38

P. Senden

884

Convergence of sin(log x)

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