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If $|a| < 1,b= \sum \limits_{k=1} ^{ \infty} \; \large\frac{a^k}{k}$ then a is equal to :

\[\begin {array} {1 1} (a)\;\sum \limits_{k=1}^{\infty} \frac{(-1)^k b^k}{k} & \quad (b)\;\sum \limits_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{k !} \\ (c)\;\sum \limits_{k=1}^{\infty} \frac{(-1)^k b^k}{(k-1)!} & \quad (d)\;\sum \limits_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{(k+1)!} \end {array}\]

1 Answer

$(b)\;\sum \limits_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{k !}$
answered Feb 27, 2014 by meena.p
 
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