If a<1,b=∑k=1∞akk⇒a=

b=∑k=1∞ akkb=a+a22+a33+.......∞ =-log (1-a)e-b=1-aa=1-e-b =1-1-b+b22-b33..... =∑k=1∞(-1)k-1bkk!

If a<1,b=∑k=1∞akk⇒a=

b=∑k=1∞ akkb=a+a22+a33+.......∞ =-log (1-a)e-b=1-aa=1-e-b =1-1-b+b22-b33..... =∑k=1∞(-1)k-1bkk!

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If $|a| < 1, b = \sum_{k = 1}^{\infty} \frac{a^{k}}{k} \Rightarrow a =$

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$\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1} b^{k}}{k!}$

$\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k} b^{k}}{k}$

$\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k} b^{k}}{(k - 1)!}$

$\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1} b^{k}}{(k + 1)!}$

answer is B.

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Detailed Solution

$\begin{array}{rcl} b & = & \sum_{k = 1}^{\infty} \frac{a^{k}}{k} \\ b & = & a + \frac{a^{2}}{2} + \frac{a^{3}}{3} + . . . . . . . \infty \\ = & - \log (1 - a) \\ e^{- b} & = & 1 - a \\ a & = & 1 - e^{- b} \\ = & 1 - (1 - b + \frac{b^{2}}{2} - \frac{b^{3}}{3} . . . . .) \\ = & \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1} b^{k}}{k!} \end{array}$