Question

The value of \[\dfrac{2}{{1!}} + \dfrac{4}{{3!}} + \dfrac{6}{{5!}} + .....\infty \] is
A. \[e\]
B. \[ - e\]
C. \[\dfrac{1}{e}\]
D. None of these

Answer 1

Hint: A series is given to us which is \[\dfrac{2}{{1!}} + \dfrac{4}{{3!}} + \dfrac{6}{{5!}} + .....\infty \]. We have to find its value. We will first find the general form of the series and then find its value using factorial and summation properties.

Formula Used: Following formulas will be useful for solving this question
\[
  a! = a \times (a - 1) \times (a - 2) \times .....\infty \\
  a! = a \times (a - 1)! \\
 \]
\[e = 1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}}.....\infty \]

Complete Step by step Solution:
 A series is given to us which is \[\dfrac{2}{{1!}} + \dfrac{4}{{3!}} + \dfrac{6}{{5!}} + .....\infty \].
We can write this series as
\[\dfrac{{2 \times 1}}{{(2 \times 1 - 1)!}} + \dfrac{{2 \times 2}}{{(2 \times 2 - 1)!}} + \dfrac{{2 \times 3}}{{(2 \times 3 - 1)!}} + .....\infty \]
Through this we can write the general form of series as \[\dfrac{{2n}}{{(2n - 1)!}}\]
Let us denote this general form with \[a\],
Therefore \[a = \dfrac{{2n}}{{(2n - 1)!}}\]
On further simplifying, we get
\[
  a = \dfrac{{2n - 1 + 1}}{{(2n - 1)!}} \\
   = \dfrac{{2n - 1}}{{(2n - 1)!}} + \dfrac{1}{{(2n - 1)!}} \\
 \]
On using \[a! = a \times (a - 1)!\], we get
\[
  a = \dfrac{{2n - 1}}{{(2n - 1)(2n - 2)!}} + \dfrac{1}{{(2n - 1)!}} \\
   = \dfrac{1}{{(2n - 2)!}} + \dfrac{1}{{(2n - 1)!}} \\
 \]
Thus, our series becomes
\[\dfrac{2}{{1!}} + \dfrac{4}{{3!}} + \dfrac{6}{{5!}} + .....\infty = \sum\limits_{n = 1}^\infty {\dfrac{1}{{(2n - 2)!}} + \dfrac{1}{{(2n - 1)!}}} \]
On further solving, we get
\[
  \dfrac{2}{{1!}} + \dfrac{4}{{3!}} + \dfrac{6}{{5!}} + .....\infty = \sum\limits_{n = 1}^\infty {\dfrac{1}{{(2n - 2)!}} + \dfrac{1}{{(2n - 1)!}}} \\
   = \dfrac{1}{{0!}} + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + .....\infty \\
   = 1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}}.....\infty \\
 \]
We know that the above series is represented by \[e\] i.e., \[e = 1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}}.....\infty \]
So, \[\dfrac{2}{{1!}} + \dfrac{4}{{3!}} + \dfrac{6}{{5!}} + .....\infty = e\]

Thus the correct option is A.

Additional Information : In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely numerous amounts, one after the other, to a given starting volume. The study of series is a major part of math and its conception and fine analysis.

Note : Students generally make mistakes while calculating factorials, generally they write\[a! = a \times (a - 2) \times .....\infty \], which is wrong. They should write and use \[ a! = a \times (a - 1) \times (a - 2) \times .....\infty = a \times (a - 1)! \]; this formula only. They also may make mistakes while determining the general form of the series. In order to determine the general form, they should follow the above steps which are done at the beginning of the “Complete step-by-step Solution” portion.