# Analyse convergence of: \$displaystyle sum^{infty}_{n=3} left (sqrt[3]{n^3 + 2} – sqrt{n^2 + 7}right)\$

If you are able to prove that both series$$sum_{n=3}^inftysqrt[3]{n^3+2}-nquadtext{and}quadsum_{n=3}^inftysqrt{n^2+7}-ntag1$$converge, or that one of them converges and the other one diverges, then you’re done.

Now, note thatbegin{align}sqrt[3]{n^3+2}-n&=sqrt[3]{n^3+2}-sqrt[3]{n^3}\&=frac2{sqrt[3]{n^3+2}^2+sqrt[3]{n^3+2}sqrt[3]{n^3}+sqrt[3]{n^3}^2}\&=frac2{sqrt[3]{n^3+2}^2+sqrt[3]{n^3+2},n+n^2}end{align}and therefore$$lim_{ntoinfty}frac{sqrt[3]{n^3+2}-n}{frac1{n^2}}=frac23$$and therefore the first series from $$(1)$$ converges.

Can you take it from here?

Hint Use that when $$alphanot = 0$$,
$$begin{equation} (1 + x)^alpha – 1 sim alpha x end{equation}$$
when $$xlongrightarrow 0$$.