Given the sum $$\sum_{n=1}^x \sin^2(\frac{33}{n})$$ is there an approximation for this sum? Preferably, it gets closer to the actual sum as $x$ grows larger (like the approximation for the harmonic numbers).
If it helps anyone, these is a closed form for $$\sum_{n=1}^x \sin^2(n)= \frac{1}{4} (2 x - \csc(1) \sin(2 x + 1) + 1)$$
Let us transform square of sine using $\sin^2 t=(1-\cos 2t)/2$ and using the Taylor expansion of cosine:
\begin{align*} \sum_{n=1}^x\sin^2\left(\frac{33}n\right)&= \frac x2-\frac 12\sum_{n=1}^x\cos\left(\frac{66}n\right) =\frac x2-\frac 12\sum_{n=1}^x\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\cdot\left(\frac{66}n\right)^{2k}\\ &=\frac x2-\frac 12\sum_{k=0}^\infty\frac{(-4356)^k}{(2k)!}\cdot H_x^{(2k)} \end{align*} where $H_x^{(2k)}$ is the generalized harmonic number.
I have got the following additional idea:
\begin{align*} \sum_{n=1}^x\sin^2\left(\frac{33}n\right) &=\frac x2-\frac 12\sum_{n=1}^x\cos\left(\frac{66}n\right) =\frac x2-\frac 12\sum_{n=1}^x\Re\left(\cos\frac 1n+\mathrm i\cdot\sin\frac 1n\right)^{66}\\[12pt] &=\frac x2-\frac 12\sum_{n=1}^x\Re\left(\sum_{k=0}^{66}\binom{66}{k}\cos^k\frac 1n\cdot\mathrm i^{66-k}\cdot\sin^{66-k}\frac 1n\right)\\[12pt] &=\frac x2-\frac 12\sum_{n=1}^x\Re\left(\sum_{k=0}^{66}\binom{66}{k}\cos^k\frac 1n\cdot\mathrm i^{66-k}\cdot\sin^{66-k}\frac 1n\right)\\[12pt] &=\frac x2-\frac 12\sum_{n=1}^x\sum_{\substack{k=0\\\text{$k$ is even}}}^{66}\binom{66}{k}(-1)^{\frac k2+1}\cos^k\frac 1n\cdot\sin^{66-k}\frac 1n\\[12pt] &=\frac x2-\frac 12\sum_{n=1}^x\sum_{j=0}^{33}\binom{66}{2j}(-1)^{j+1}\cos^{2j}\frac 1n\cdot\sin^{66-2j}\frac 1n\\[12pt] &\approx\frac x2-\frac 12\sum_{n=1}^x\sum_{j=0}^{33}\binom{66}{2j}(-1)^{j+1}\left(1-\frac 1{2!\cdot n^2}+\frac 1{4!\cdot n^4}-\frac 1{6!\cdot n^6}\right)^{2j}\\[12pt] &\quad\cdot\left(\frac 1n-\frac 1{3!\cdot n^3}+\frac 1{5!\cdot n^5}-\frac 1{7!\cdot n^7}\right)^{66-2j} \end{align*}
The more terms of Taylor series of sine and cosine function are used the better approximation you get.