$\mathop \sum \limits_{k = 1}^\infty \frac{1}{{k!}}\left( {\mathop \sum \limits_{n = 1}^\infty {2^{n - 1}}} \right)$ is equal to

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Analyzing +4 | -1

\sum \limits_{k = 1}^\infty \frac{1}{{k!}}\left( {\sum \limits_{n = 1}^\infty {2^{n - 1}}} \right)

is equal to

View answer and explanation

Correct option: D

{e^2} - e

Option D (

{e^2} - e

) is correct because it is derived by splitting the sum into two separate exponential series and evaluating them.

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