# Math Help - Test for convergence...

1. ## Test for convergence...

Test the series for convergence or divergence.

The sum figure from K=1 to infinity 2 / ((k^2)*(1+k^2))^(1/3)

2. $\frac{1}{\sqrt[3]{k^{2}\left( 1+k^{2} \right)}}<\frac{1}{k^{4/3}},$ so what we conclude?

3. Convergent. Thanks! Why did you use 4/3?

4. from what i remember if its goes to zero faster then 1/x it is convergencing
and it does

(dont have the proof with me)

5. ah, it's quite simple.

since it's for each $k\ge1,$ then it's obvious that $\frac{1}{k^{2}\left( 1+k^{2} \right)}<\frac{1}{k^{2}\cdot k^{2}}=\frac{1}{k^{4}},$ and you'll get the inequality.