$Let \ \dfrac {x}{e^x-1}= \sum^{\infty}_{i=0}a_ix^i $ Where ais are the co-efficient of the corresponding exponent of x. $x=(e^x-1)(a_0+a_1x+a_2x^2+a_3x^3+ \cdots )$ $x= \left ( 1+ x+ \dfrac {x^2}{2!}+ \dfrac {x^3}{3!}+ \cdots \ cdots -1 \right ) ( a_0 + a_1 x+a_2x^2 +a_3 x^3 + \cdots \ \cdots ) $ $ x=\left ( x+ \dfrac {x^2}{2}+ \dfrac {x^3}{6}+ \dfrac {x^4}{24}+ \dfrac {x^5}{120} \ \cdots \right ) (a_0+a_1x+a_2x^2 + a_3x^3+a_4x^4 \ \cdots \ \cdots) $ $x=a_0x+a_1x^2+a_2x^3+a_3x^4 +a_4x^5 \cdots + \dfrac {a_0x^2}{2}+ \dfrac {a_1x^3}{2}+ \dfrac {a_2x^4}{2}+ \dfrac {a_3 x^5}{2} \cdots + \dfrac {a_0 x^3}{6} + \dfrac {a_1x^4}{6}+ \dfrac {a_2x^5}{6}+ \cdots + \dfrac {a_0 x^4}{24}+ \dfrac {a_1x^5}{24}\cdots + \dfrac {a_0x^5}{120}+ \cdots$ Comparing the co-efficients of x on both sides, $1=a_0$ Comparing the co-efficients of x2 on both sides, $0=a_1+ \dfrac {a_0}{2}$ $a_1= -\dfrac {1}{2}$ Comparing the co-efficients of x3 on both sides, $0=a_2+ \dfrac {a_1}{2}+ \dfrac {a_0}{6} \ a_2 = \dfrac {1}{4}- \dfrac {1}{6} \ a_2 = \dfrac {1}{12} $ Comparing the co-efficients of x4 on both sides, $0=a_3+\dfrac {a_2}{2}+ \dfrac {a_1}{6}+ \dfrac {a_0}{24} \ a_3 = - \dfrac {1}{24}+ \dfrac {1}{12}- \dfrac {1}{24}= \dfrac {-1+2-1}{24} \ a_3=0 $ Comparing the co-efficients of x5 on both sides, $0=a_4+\dfrac {a_3}{2}+ \dfrac {a_2}{6}+ \dfrac {a_1}{24}+ \dfrac {a_0}{120} \ a_4 = 0 - \dfrac {1}{72}+ \dfrac {1}{48}- \dfrac {1}{120} \ a_4 = -\dfrac {1}{720} $ $\therefore \dfrac {x}{e^x-1}= \sum^{\infty}_{i=0}a_ix^i \ \dfrac {x}{e^x-1}=1 - \dfrac {x}{2}+ \dfrac {x^2}{12}- \dfrac {x^4}{720}+ \cdots $ $Adding \ \dfrac {x}{2} \ on \ both \ sides,$ $x\left [ \dfrac {1}{e^x-1}+ \dfrac {1}{2} \right ]= 1+ \dfrac {x^2}{12}- \dfrac {x^4}{720}+ \cdots \ \dfrac {x}{2} \left [ \dfrac {e^x+1}{e^x-1} \right ]=1 + \dfrac {1}{12}x^2 - \dfrac {1}{720}x^4 + \cdots $