e $^x \; = \;1+ x+ \dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\ldots \;\;\;\;\;\;\;\; Standard \; Expansion \\ \; \\ \; \\ \therefore \dfrac{x}{e^x-1} \; = \; \dfrac{x}{\bigg[1+ x+ \dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\ldots \bigg] \; - \;1} \\ \; \\ = \dfrac{x}{ x \; \bigg[ 1+ \dfrac{x}{2!}+\dfrac{x^2}{3!}+\dfrac{x^3}{4!}+\ldots \bigg] } \\ \; \\ = \bigg[ 1+ \dfrac{x}{2!}+\dfrac{x^2}{3!}+\dfrac{x^3}{4!}+\ldots \bigg]^{-1} \\ \; \\ = \; 1 \; - \; \bigg(\dfrac{x}{2} +\dfrac{x^2}{6}+\dfrac{x^3}{24}+\dfrac{x^4}{12}+\ldots \bigg) \; + \; \bigg(\dfrac{x}{2} +\dfrac{x^2}{6}+\dfrac{x^3}{24}+\dfrac{x^4}{12}+\ldots \bigg)^2 \\ \; - \; \bigg(\dfrac{x}{2}+ \dfrac{x^2}{6}+\dfrac{x^3}{24}+\dfrac{x^4}{12}+\ldots \bigg)^3 \; + \; \bigg(\dfrac{x}{2}+ \dfrac{x^2}{6}+\dfrac{x^3}{24}+\dfrac{x^4}{12}+\ldots \bigg)^4 \\ \; \\ \; \\ = \; 1-\dfrac{x}{2} \; + \; x^2 \bigg( -\dfrac{1}{6} + \dfrac{1}{4} \bigg) \; + \; x^3 \bigg( -\dfrac{1}{24} + \dfrac{1}{6} -\dfrac{1}{8} \bigg) \; + \; \\ x^4 \bigg( -\dfrac{1}{120} + \dfrac{1}{36} +\dfrac{1}{24} -\dfrac{1}{8} + \dfrac{1}{16} \bigg)+ \ldots \\ \; \\ \; \\ \; \\ =1-\dfrac{x}{2}+\dfrac{x^2}{12}+x^3(0) + \bigg( \dfrac{-6+20+30-90+45}{720} \bigg)x^4+\ldots \\ \; \\ \therefore \dfrac{x}{e^x-1} \; = \; 1- \dfrac{x}{2}+\dfrac{x^2}{12}-\dfrac{x^4}{720}+\ldots \\ \; \\ \; \\ \frac{x}{2} \bigg[ \dfrac{e^x+1}{e^x-1} \bigg] \; = \; \frac{x}{2} \bigg[ 1\;+\; \dfrac{2}{e^x-1} \bigg] \\ \; \\ = \dfrac{x}{2} \; + \; \dfrac{x}{e^x-1} \\ \; \\ = \dfrac{x}{2} \; + \; 1- \dfrac{x}{2}+\dfrac{x^2}{12}-\dfrac{x^4}{720}+\ldots \\ \; \\ = 1- \dfrac{x^2}{12}-\dfrac{x^4}{720}+\ldots \\ \; \\ \; \\ \frac{x}{2} \bigg[ \dfrac{e^x+1}{e^x-1} \bigg] \; = \; 1 + \dfrac{1}{12}x^2-\dfrac{1}{720}x^4+\ldots \;\;\;\;\; \ldots Hence \; Proved. $