I think yes, since the first would be a kind of subsequence of the partial sums of $\dfrac{1}{k^2}$...
To provide some context, the question arised while studying Fourier Series on Apostol, when was stated that if $\{\phi_n\}$ is a orthonormal sequence of functions in some interval [a,b],and f is Riemman integrable there, then we have $\sum_{k=1}^\infty {c_k}^2 \le \int_a^b f^2(t) \, dt$, where the $c_k's$ are given by:
$$c_k=\int_a^b f(t) \phi_k(t) \, dt.$$
Taken specifically the $\phi_n(t)$ as $\dfrac{1}{\sqrt{2\pi}}, \dfrac{\cos t}{\sqrt{\pi}},\dfrac{\sin t}{\sqrt{\pi}},\dfrac{\cos 2t}{\sqrt{\pi}}, \dfrac{\sin 2t}{\sqrt{\pi}}$,etc., we get the very well known fourier series of f(t), but when trying to applicate the inequality stated above, I get into trouble, since the book states it should be $\dfrac{{a_o}^2}{2}+\sum_{k=1}^\infty a_k^2+b_k^2 \le \dfrac 1 \pi \int_{-\pi}^\pi f^2(x)\,dx $, but my computations show that $c_2^2=\pi a_1^2, c_3^2=\pi b_1^2, c_4^2=\pi a_2^2$, etc., and not, for example $c_2= \pi(a_1^2+b_1^2), c_3=\pi(a_2^2+b_2^2)$, but it could be interpreted as a subsequence of the previous by taking odd partial sums...Am I correct?
No: the series is absolutely convergent, so
$$\sum_{k\ge 1}\left(\frac1{k^2}+\frac1{(k+1)^2}\right)=1+2\sum_{k\ge 2}\frac1{k^2}=2\sum_{k\ge 1}\frac1{k^2}-1=\frac{\pi^2}3-1\ne\frac{\pi^2}6=\sum_{k\ge 1}\frac1{k^2}\;.$$