From Fourier Analysis an Introduction pg 87:
"Finally, we need to find a convergent series of positive terms $\sum \alpha_k$ and a sequence of integers {$N_{k}$} which increases rapidly enough so that: $$(i)N_{k+1}>3N_{k}$$ $$(ii)\alpha_{k}\log{N_{k}}\to \infty\ \ as\ k\to\infty$$ We choose (for example) $\alpha_{k}=\frac{1}{k^{2}}$ and $N_{k}=3^{2^k}$ which are easily seen to satisfy the above criteria."
I have doubted it if it really satisfies the second condition, Because: $$\lim_{k\to\infty}\frac{\log{3^{2^{k}}}}{k^2}=\frac{\log{9^{k}}}{k^2}=\frac{k\log{9}}{k^2}=\frac{\log{9}}{k}\to 0$$ or $$\lim_{k\to\infty}\frac{\log{3^{2^{k}}}}{k^2}=\frac{\log{3^{k^{2}}}}{k^2}=\frac{k^{2}\log{3}}{k^2}=\log{3}$$ I don't get the example above, it doesn't satisfy the (ii) condition. Can someone help me with it?
Unfortunately, you got it wrong both times. The sequence $N_k=3^{2^k}$ means that you raise 3 to the power of $2^k$, so you can write $N_k = 3^{(2^k)}$.
As for the limit, it does indeed satisfy the conditions because
$$\lim\limits_{k\to\infty} \frac{\log3^{2^k}}{k^2} = \lim\limits_{k\to\infty} \frac{2^k\log3}{k^2} = +\infty$$