<s>[INST] Convergence of series $\sum_{k=1}^\infty (\frac{1+k^2}{1+k^3})^2$ Given the series
$$
\sum_{k=1}^\infty (\frac{1+k^2}{1+k^3})^2,
$$
can I conclude from
$$
\begin{align}
\sum_{k=1}^\infty (\frac{1+k^2}{1+k^3})^2
&=\sum_{k=1}^\infty (\frac{1+k^2}{1+k^3})(\frac{1+k^2}{1+k^3})\\
&\leq\sum_{k=1}^\infty (\frac{k^2+k^2}{1+k^3})(\frac{k^2+k^2}{1+k^3})\\
&\leq\sum_{k=1}^\infty (\frac{2k^2}{k^3})(\frac{2k^2}{k^3})\\
&=\sum_{k=1}^\infty (\frac{2}{k})(\frac{2}{k})\\
&=\sum_{k=1}^\infty (\frac{2}{k^2}),
\end{align}
$$
that  $\sum_{k=1}^\infty (\frac{1+k^2}{1+k^3})^2$ must converge since  $\sum_{k=1}^\infty (\frac{2}{k^2})$ converges and the above estimations are valid?
 [/INST] Yes, it is correct, except that after the final $=$ sign you should get $\sum_{k=1}^\infty\frac4{k^2}$.
Besides, note that the inequality$$\frac{1+k^2}{1+k^3}\leqslant\frac2k\tag1$$is equivalent to $k+k^3\leqslant2+2k^3$, which, in turn, is equivalent to $k\leqslant2+k^3$, which is true, since you actually have $k\leqslant k^3$ for each natural $k$. So, this gives you another way of proving $(1)$.
</s>