Examine the convergence of the following series:
$$ a) \qquad \sum_{k=0}^{\infty} \Bigg(\frac{k^2+1}{(k+1)^2}\Bigg)^2 ; $$
$$ b) \qquad \sum_{k=2}^{\infty} \frac{2^k}{\lfloor\frac{k}{2}\rfloor!} $$
In part b), the floor function $ \lfloor a \rfloor$ function returns the largest integer less than or equal to a given number.
As I'm really stuck on these two, can you give me any ideas/solutions/hints?
The first series is not convergent since the general term tends to 1 in infinity. For the second one, put $U_k:=\frac{2^k}{[k/2]}.$ We have $$\lim_{k\to \infty} \frac{U_{k+1}}{U_k}= \lim_{k\to \infty} \frac{2}{[(k+1)/2]}=0<1$$ then by d'Alembet criterion, we get the convergence of the second series. (for the d'Alembert criterion, see https://fr.wikipedia.org/wiki/R%C3%A8gle_de_d'Alembert)