Sum of series converge or diverge

Question

does the sum of ∑ (from k = 1 to infinity), of [e^{sin (k)}]/k converge or diverge

Answer 1

This series is divergent

Since sin⁡(k) oscillates between -1 and 1 for all k, the values of e^{\sin(k) will oscillate between e⁻¹≈0.3679 ande^1≈2.718.

Therefore, e^{\sin(k)} is bounded by two constants: 0.3679 ≤e^(sin⁡(k))≤2.718

To use the comparison test, consider a simpler series ∑C/k where C is a constant. Since e^{\sin(k)} is bounded, we can use these bounds for comparison.

We know the harmonic series ∑1/k diverges. So, comparing ∑e*(sin⁡(k))/k to the harmonic series:

0.3679⋅∑1/k ≤ ∑e^(sin⁡(k))/k≤2.718⋅∑1/k

Since ∑1/k diverges, the series ∑e^(sin⁡(k))/k also diverges by comparison to the harmonic series.