Acceleration of trigonometric series convergence

Question

Using Cauchy Theorem I figured out some formulas for accelerating the convergence of trigonometric series.

Let $f$ be a rational function, $\lambda \in (0, \frac{1}{2}]$, then $\sum\limits_{\substack{n=1}}^\infty {f}(n)e^{2\pi i \lambda n} = C_f(\lambda) + \frac{i}{2\sqrt{\lambda}}\sum\limits_{n=1}^\infty \frac{{f}\left(\frac{n i}{\sqrt{\lambda}}\right)e^{-\frac{(2\lambda-1)\pi n}{\sqrt{\lambda}}}-{f}\left(-\frac{n i}{\sqrt{\lambda}}\right)e^{\frac{(2\lambda-1)\pi n}{\sqrt{\lambda}}}}{\sinh{\frac{\pi n}{\sqrt{\lambda}}}}-\sum\limits_{n=1}^\infty \frac{{f}(n)e^{2\pi i \lambda n} - {f}(-n)e^{-2\pi i \lambda n}}{e^{2\pi n \sqrt{\lambda}}-1}$
where $C_f(\lambda)$ is some closed form expansion.

As a corollary (for $\lambda = \frac{1}{2}$) we have a formula for alternating series $$\sum\limits_{n=1}^\infty (-1)^n{f}(n) = C_f + \frac{i}{\sqrt{2}}\sum\limits_{n=1}^\infty \frac{{f}(\sqrt{2}ni)-{f}(-\sqrt{2}ni)}{\sinh(\sqrt{2}\pi n)}-\sum\limits_{n=1}^\infty\frac{(-1)^n({f}(n)-{f}(-n))}{e^{\sqrt{2}\pi n}-1}$$

Are these formulas well-known? They look quite similar to ones by Ramanujan or Abel-Plana, but I couldn't find exactly the same results.