Evaluating $\lim_{k\to+\infty}\frac12\sin(\sqrt{k+1})-2 \left(\sin(\sqrt{k+1})-\sqrt{k+1}\cos(\sqrt{k+1})\right)+\sum_{n=0}^k\sin(\sqrt n)$

Question

I need help at evaluating this to some closed form formula: $$\lim_{k\to+\infty}\frac{\sin\left(\sqrt{k+1}\right)}{2}-2 \left(\sin\left(\sqrt{k+1}\right)-\sqrt{k+1}\cos\left(\sqrt{k+1}\right) \right)+\sum_{n=0}^k \sin \left(\sqrt n\right)$$

I started with a C++ code to get an approximation for this limit, and after trying multiple techniques this is the best approximation: $$-0.203568...$$ and it's suspiciously looks like it equals to this integral that i have evaluted in WolframAlpha: $$-2\int_{0}^{\infty}\frac{\cos\left(\sqrt{\frac{x}{2}}\right)\sinh\left(\sqrt{\frac{x}{2}}\right)}{e^{2\pi x}-1}dx = -0.20356860652805711756342....$$ and we can totally notice that my approximation is identical with this integral for 6 decimal digits

My questions:
is there's a way to actually prove or disprove this relation?
and is it possible to give this limit a closed form if it's not actually equal to this integral?
complicated techniques are welcome to me.

Answer 1

This is not an answer.

The integral is interesting if you write it as $$\int_{0}^{\infty}\frac{\cos\left(\sqrt{\frac{x}{2}}\right)\sinh\left(\sqrt{\frac{x}{2}}\right)}{e^{2\pi x}-1}\,dx=\sum_{n=0}^\infty\int_{0}^{\infty}\frac{ \cos\left(\sqrt{\frac{x}{2}}\right)\sinh\left(\sqrt{\frac{x}{2}}\right)}{e^{2 \pi (n+1) x} }\,dx$$ $$J_n=\int_{0}^{\infty}\frac{ \cos\left(\sqrt{\frac{x}{2}}\right)\sinh\left(\sqrt{\frac{x}{2}}\right)}{e^{2 \pi (n+1) x} }\,dx$$

$$J_n=\frac{\cos \left(\frac{1}{8 \pi (n+1)}\right)-\sin \left(\frac{1}{8 \pi (n+1)}\right)}{8 \pi (n+1)^{3/2}}$$

Let $\frac{1}{8 \pi (n+1)}=x_n$ to make $$J_n=2 \sqrt{2 \pi }\, x_n^{3/2}\,\, \big(\cos (x_n)-\sin (x_n)\big)$$

The summation converges because $$J_n=\frac{1}{8 \pi n^{3/2}}\Bigg(1-\frac{12 \pi+1 }{8 \pi n}+\frac{240 \pi ^2+40\pi-1}{128 \pi ^2 n^2}+O\left(\frac{1}{n^3}\right) \Bigg)$$ but quite slowly because $$\frac {J_{n+1}}{J_n}=1-\frac{3}{2 n}+\frac{27 \pi+1 }{8 \pi n^2}+O\left(\frac{1}{n^3}\right)$$

Expanded again and summing, we have $$\sum_{m=0}^\infty \left(\sin \left(\frac{(m+1)\pi}{2} \right)+\cos \left(\frac{(m+1)\pi}{2} \right)\right)\, \frac {\zeta \left(m+\frac{3}{2}\right)}{(8\pi)^{m+1}\, m! }$$ which converges very fast.

Answer 2

If we use Euler-MacLaurin summation, we have $$\sum_{n=1}^k \sin \left(\sqrt{n}\right)=C+\Bigg(\frac{5}{2}-\frac{1}{1920 k^2}-\frac{1}{96768 k^3}+O\left(\frac{1}{k^4}\right)\Bigg) \sin \left(\sqrt{k}\right)-$$

$$2\sqrt k\Bigg(1-\frac{1}{48 k}-\frac{1}{11520 k^2}+\frac{503}{1935360 k^3}+O\left(\frac{1}{k^4}\right)\Bigg) \cos \left(\sqrt{k}\right)$$

For this level of expansion

$$C=\frac{5245636079431349 \cos (1)-4015687077315245 \sin (1)}{2678117105664000}=-0.203447$$