Take a look at the averaging sum



> 
> $$\frac{\pi}{n}\sum\_{k=1}^n\;\exp{(-\sin\theta\_k)}\cdot \sin(\theta\_k +\cos\theta\_k)\, \qquad\text{where }\;\theta\_k=(2k-1)\frac{\pi}{2n}$$
> 
> 
> 


depending on $n\in\mathbb{N}$.


**How could one analyse convergence for $n\rightarrow\infty$, and possibly compute its limit? ...** 


**but 'ignoring' the fact that it is equal to the point evaluation of the Cosine transform**
$$\int\_{-\infty}^\infty\; \frac{\cos(\omega x)}{x^{2n}+1}\, dx$$
for $\omega = 1$, cf the [Math. SE answer](https://math.stackexchange.com/a/1785873/316749). If $n\to\infty$ then $1/(x^{2n}+1)$ converges pointwise to the characteristic function $\chi\_{[-1,1]}$ (ignoring $x=\pm 1$). Hence the sought limit should equal
$$\int^1\_{-1}\cos x\, dx = 2\sin(1)\approx 1.6829$$


Let me repeat my question as "How to tame the above expression when encountered in the wild?"  

Thus deliberately excluding the ‘Fourier background’, I'd like to know appropriate methods to determine the limit.  

The above formulation *"limit should equal"* could then be replaced with the statement *"is convergent with limit equal to".*



Methods to tackle this series and get to the limit?
Fodor's hint works.
$$
\frac{\pi}{n}\sum\_{k=1}^n\;\exp{(-\sin\theta\_k)}\cdot \sin(\theta\_k +\cos\theta\_k)\, \qquad\text{where }\;\theta\_k=(2k-1)\frac{\pi}{2n}
$$
is a Riemann sum for the integral
$$
\int\_0^\pi \exp{(-\sin t)}\cdot \sin(t +\cos t)\;dt = 2\sin(1)
$$
This integral is easy because of the antiderivative
$$
\int \exp{(-\sin t)}\cdot \sin(t +\cos t)\;dt = -e^{-\sin t}\sin(\cos t) + C
$$


