# real analysis – Need help on the general formula for the Taylor series of \$frac{2x}{e^{2x}-1}\$

I need help on deriving the general formula for the Taylor series of $$frac{2x}{e^{2x}-1}$$

It is known that the series for $$dfrac{x}{e^{x}-1}$$ is: $$dfrac{x}{e^{x}-1}=sum_{n=0}^{+infty}dfrac{B_nx^{n}}{n!}$$

So the question is whether the series for $$frac{2x}{e^{2x}-1}$$ is:

$$frac{2x}{e^{2x}-1}=sum_{n=0}^{+infty}dfrac{B_n(2x)^{n}}{n!}tag{1}$$

or

$$frac{2x}{e^{2x}-1}=2left(dfrac{x}{e^{2x}-1}right)=2left(sum_{n=0}^{+infty}dfrac{B_n(2x)^{n}}{n!}right)tag{2}$$

So which series is the correct representation for $$frac{2x}{e^{2x}-1}$$? $$(1)$$ or $$(2)$$

I suspect that it must be $$(2)$$, because if it is $$(1)$$, how does one represent the series for $$dfrac{3x}{e^{2x}-1}$$