Question

Question 13
Test the following series for convergence or divergence.
È sin(1/n)
-1​

Answer 1

Step-by-step explanation:

The series Sigma [sin(1/n)] is a divergent series, because lim(n-->inf.)[sin(1/n)]/(1/n) = 1, which is nonzero, which in turn is a consequence of lim(x-->0)[sin(x)]/x = 1. Hence the two series Sigma [sin(1/n)] and Sigma(1/n) have the same convergence behaviour by limit comparison test, for series of positive terms But we know that the harmonic series Sigma(1/n) diverges. Hence the given series Sigma[sin(1/n) also diverges.

Note that for all 1 </= n, 0 < (1/n) </= 1 < π, and so sin(1/n) is positive i.e. the given series is of positive terms.

Answer 2

Answer:

Since

limn→∞sin(1n)1n=1

so

sin(1n)∼∞1n

so the series ∑n≥1sin(1n) is divergent by comparison with the series ∑n≥11n.

We have:

∀x∈[0,π2]:2xπ≤sinx

Hence:

2πn≤sin1n

Since ∑1/n diverges, ∑sin1/n diverges too.