Sum of series of the form $sum frac{(-1)^{m+1}}{2m+1}sin(frac{pi}{2}(2m+1)x)$

I would like to calculate the sum of the series:

$$\begin{equation} \sum_{m=M+1}^{\infty}\frac{(-1)^{m+1}}{2m+1}\sin((2m+1)\frac{\pi}{2}x) \end{equation}$$
where M is big and finite.

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I searched on the books and found this sum:
$$\begin{equation} \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}\sin((2k-1)x)=\frac{1}{2}\ln\tan(\frac{\pi}{4}+\frac{x}{2}) \end{equation}$$
Now I try to put my question in the above form. Let m-M=n and m=n+M
$$\begin{equation} \sum_{n=1}^{\infty} \frac{(-1)^{n+M+1}}{2(n+M)+1}\sin(2(n+M)+1)\frac{\pi}{2}x) \end{equation}$$
I do not know how to proceed further.