Testing Convergence: Convergence Tests

Question

I need help finding wether the following series converges or diverges. Thank you!

∑𝑘=1 ∞. 𝑘 sin(𝑘) / (𝑘^2 + 1)

Answer 1

All sums in this problem are k=1 to infinity

The sum in this problem may be well known, but I will show convergence by using the fact that ∑ sin(k)/k = (π- 1)/2

For a demonstration of this sum, see https://sumantmath.wordpress.com/2019/11/25/sum-of-series-sink-k-using-fourier-series/

Note that 1/k - k/(k^2+1) = (k^2+1)/((k(k^2+1)) - k*k/(k(k^2+1)) = 1/(k(k^2+1))

Then, for the above series, ∑k sin(k)/(k^2+1) = ∑ sin(k)/k - (∑ sin(k)/k - ∑k sin(k)/(k^2+1)) =

(π - 1)/2 - ∑ sin(k)/(k(k^2+1))

Thus, to show that ∑k sin(k)/(k^2+1) converges, we only must show that ∑ sin(k)/(k(k^2+1)) converges.

We can show that this sum converges absolutely by showing that its absolute value is bounded above by a series that we will show converges |sin(k)/(k(k^2+1))| < 1/(k(k^2+1)) < 1/k^3

∑1/k^3 < 1 + ∫₁^∞ 1/k³ = 1 + (-1/(2k²) |₁^∞ = 1 + (0 --1/2) = 3/2

Thus, ∑ sin(k)/(k(k^2+1)) converges absolutely, and (π - 1)/2 - ∑ sin(k)/(k(k^2+1)) converges, but this equals ∑k sin(k)/(k^2+1), so this converges.