Consider the Dirichlet kernel $D_N(x)=\sum_{|k|\le N} e^{ikx}$. Its second derivative reads as $$D_N^{\prime\prime}(x) = -\sum_{|k|\le N} e^{ikx}k^2.$$
What is the asymptotic order of $|D_N^{\prime\prime}(0)|^2$ as $N \to \infty$ ?
My attempt:
$$|D_N^{\prime\prime}(x)|^2 = \sum_{|k|\le N} |e^{ikx} |^2 k^4 + 2 \sum_{|k|\le N} \sum_{h \neq k} |e^{ikx} | |e^{ihx} | k^2 h^2 \cos (kx-hx) $$ $$= \sum_{|k|\le N} k^4 + 2 \sum_{|k|\le N} \sum_{h \neq k} k^2 h^2 \cos (kx-hx)$$
Hence, for $x=0$, $$|D_N^{\prime\prime}(0)|^2 = \sum_{|k|\le N} k^4 + 2 \sum_{|k|\le N} \sum_{h \neq k} k^2 h^2, $$ where
$$\sum_{|k|\le N} k^4 = 2 \sum_{k=1}^N k^4 = O(N^5),$$
$$\sum_{|k|\le N} \sum_{h \neq k} k^2 h^2 =4 \sum_{k=1}^N \sum_{h \neq k} k^2 h^2 + 2\sum_{k=1}^N k^4 . $$
Focus on $\sum_{k=1}^N \sum_{h \neq k} k^2 h^2$. The latter can be rewritten as
$$\sum_{k=2}^N \sum_{h=1}^{k-1} k^2 h^2+ \sum_{k=1}^{N-1} \sum_{h = k+1}^N k^2 h^2 $$ $$=\sum_{k=2}^N k^2 \sum_{h=1}^{k-1} h^2+ \sum_{k=1}^{N-1} k^2 \left(\sum_{h = 1}^N h^2 - \sum_{h = 1}^k h^2\right)$$ $$=\sum_{k=2}^N k^2 \frac{1}{6} k(k-1)(2k-1)+ \sum_{k=1}^{N-1} k^2 \frac{1}{6}\left( N(N+1)(2N+1) - k(k+1)(2k+1) \right)$$ $$ \sim \frac{1}{3} \sum_{k=2}^N k^5 + \frac{1}{3} N^3\sum_{k=1}^{N-1} k^2 - \frac{1}{3} \sum_{k=1}^{N-1} k^5 = -\frac{1}{3} +\frac{1}{3}N^5 + \frac{1}{3} N^3 \frac{1}{6}N(N-1)(2N-1) = O(N^6).$$
So, finally, $|D_N^{\prime \prime}(0)|^2 = O(N^6)$.
Is my attempt fully correct?
I am puzzled by the fact that the leading term seems to be $\sum_{|k|\le N} \sum_{h \neq k} k^2 h^2$ and not $\sum_{k=1}^N k^4$.