Test the series for convergence or divergence. The sum figure from K=1 to infinity 2 / ((k^2)*(1+k^2))^(1/3)
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$\displaystyle \frac{1}{\sqrt[3]{k^{2}\left( 1+k^{2} \right)}}<\frac{1}{k^{4/3}},$ so what we conclude?
Convergent. Thanks! Why did you use 4/3?
from what i remember if it`s goes to zero faster then 1/x it is convergencing and it does (don`t have the proof with me)
ah, it's quite simple. since it's for each $\displaystyle k\ge1,$ then it's obvious that $\displaystyle \frac{1}{k^{2}\left( 1+k^{2} \right)}<\frac{1}{k^{2}\cdot k^{2}}=\frac{1}{k^{4}},$ and you'll get the inequality.
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