Why convergence of these two series is equivalent?

Question

In my notes I have the following:

For $s\geq 0$, define the Sobolev Space $H_s$ by: $$H_s := \left\{f\in L^2([0, 2\pi], \mathbb{C})\,\, : \,\, \sum_{k=-\infty}^\infty |k|^{2s}|\hat{f_k}|^2 <\infty\right\}$$ And define the Inner Product as: $$(f, g)_{(s)} := \sum_{k=-\infty}^\infty \hat{f_k}\bar{\hat{g_k}}(1 + k^2)^s$$ Note that $\sum_{k=-\infty}^\infty (|k|^2)^s|\hat{f_k}|^2$ converges iff $\sum_{k=-\infty}^\infty (1+k^2)^s |\hat{f_k}|^2$ converges.

How do we show the last statement is true? From how the lecturer put it down it seems VERY obvious as if you could do it in one line, but I couldn't show this is true.

My Attempt

I tried as follows: Since $s\geq 0$ then clearly: $$(1 + k^2)^s \geq (|k|^2)^s \quad \forall k\in\mathbb{Z}\quad \forall s\geq 0$$ since $k^2 = |k|^2$ for all $k\in\mathbb{Z}$. Therefore we must have: $$(1+k^2)^s |\hat{f_k}|^2 \geq (|k|^2)^s |\hat{f_k}|^2\quad \forall k\in\mathbb{Z}\quad \forall s\geq 0$$ so that we can show the direction "$\Longleftarrow$" using the comparison test.

Answer 1

Hint:

Note the general terms of both series (which are positive) are equivalent sice $$\lim_{k\to\infty}\frac{(1+k^2)^s|\hat f_k|^2}{k^{2s}|\hat f_k|^2}=\lim_{k\to \infty}\Bigl(1+\frac1{k^2} \Bigr)^s=1.$$ Hence both series converge or both diverge.

Answer 2

Hint: Use that $|k|\ge 1$ for $k\in\mathbb{Z}\setminus\{0\}$, so $2k^2 \ge k^2+1$.