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

$\mathop \sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^{k - 1}}{b^k}}}{{k!}}$

$\mathop \sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^k}{b^k}}}{k}$

$\mathop \sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^k}{b^k}}}{{\left( {k - 1} \right)!}}$

$\mathop \sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^{k - 1}}{b^k}}}{{\left( {k + 1} \right)!}}$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Analyzing +4 | -1

\left| a \right| < 1,\;b = \sum \limits_{k = 1}^\infty \frac{{{a^k}}}{k}

, then

a

is equal to

View answer and explanation

Correct option: B

\sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^{k - 1}}{b^k}}}{{k!}}

Option B (

\sum \limits_{k = 1}^\infty \frac{{{{\left( { - 1} \right)}^{k - 1}}{b^k}}}{{k!}}

) is correct because it represents the Maclaurin series expansion of

a

in terms of

b

, derived from

b = -\log(1-a)