Question

The sum of the infinite series… $1 + \dfrac{1}{{2!}} + \dfrac{{1 \cdot 3}}{{4!}} + \dfrac{{1 \cdot 3 \cdot 5}}{{6!}} + \cdot \cdot \cdot$ is equal to
(1) $e$
(2) ${e^2}$
(3) $\sqrt e$
(4) $\dfrac{1}{e}$

Answer 1

We start solving the problem by making the simplifications in the given series and write the general term (or nth term) that represents every term of the series. After that we will apply summation to the nth term and the formula of sum of exponent series i.e, ${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....$ and find the sum of the infinite terms of the series.

Complete answer: Let the given series be $S$
$\Rightarrow S = 1 + \dfrac{1}{{2!}} + \dfrac{{1 \cdot 3}}{{4!}} + \dfrac{{1 \cdot 3 \cdot 5}}{{6!}} + \cdot \cdot \cdot {\text{ }} - - - - \left( 1 \right)$
We have to find the sum of the given series.
We can represent the above series by writing the general term (or nth term) of the series. As the series has no end, it means we have to find the sum upto infinite.
Now we can observe that the first term has no effect while writing the general term. So, we can skip the first term and observe the series from the second term to write its general term.
Now, we can observe that second term has $1$ in the numerator and $2!$ in the denominator. Similarly, third term has multiplication of two odd numbers i.e., $1 \cdot 3$ in the numerator and $4!$ in the denominator and hence, series goes on i.e., numerator has sequence of $1,{\text{ }}1 \cdot 3,{\text{ }}1 \cdot 3 \cdot 5,{\text{ }}.....$ and denominator has sequence of $2!,{\text{ }}4!,{\text{ }}6!,{\text{ }}....$
Let denote the nth term by ${T_n}$
So, the nth term can be written as
${T_n} = \dfrac{{\left( {1 \cdot 3 \cdot 5 \cdot 7.....(2n - 1)} \right)}}{{2n!}}{\text{ }} - - - \left( 2 \right)$
Because in the numerator we have multiplication of odd terms and in denominator we have term $2n!$ in the denominator
In equation $\left( 2 \right)$ if we observe, we will find that all the even terms are missing in the numerator
So, we will multiply and divide by even terms i.e., $\left( {2 \cdot 4 \cdot 6 \cdot 8....2n} \right)$ in equation $\left( 2 \right)$ to simplify the nth term.
$\therefore$ equation $\left( 2 \right)$ becomes,
${T_n} = \dfrac{{\left( {1 \cdot 3 \cdot 5 \cdot 7.....(2n - 1)} \right)}}{{2n!}} \times \dfrac{{\left( {2 \cdot 4 \cdot 6.....2n} \right)}}{{\left( {2 \cdot 4 \cdot 6.....2n} \right)}}{\text{ }} - - - \left( 3 \right)$
We know that $n! = n\left( {n - 1} \right)\left( {n - 2} \right).......1$
So, if we observe in the numerator, we get the $2n!$
$\therefore$ equation $\left( 3 \right)$ becomes,
${T_n} = \dfrac{{2n!}}{{2n!\left( {2 \cdot 4 \cdot 6....2n} \right)}}$
$\Rightarrow {T_n} = \dfrac{1}{{\left( {2 \cdot 4 \cdot 6....2n} \right)}}{\text{ }} - - - \left( 4 \right)$
Now if we take $2$ common from the denominator, then it can be written as ${2^n} \times \left( {1 \cdot 2 \cdot 3.....\left( {n - 2} \right)\left( {n - 1} \right)n} \right){\text{ }} = {2^n} \times n!$
So, equation $\left( 4 \right)$ becomes,
${T_n} = \dfrac{1}{{{2^n} \times n!}}$
Apply summation to the ${T_n}$ and substitute the value of ${T_n}$ in equation $\left( 1 \right)$ we get
$S = 1 + \sum {T_n}$
$\Rightarrow S = 1 + \sum \dfrac{1}{{{2^n} \times n!}}$
This can be written as,
$\Rightarrow S = 1 + \dfrac{1}{{2\left( {1!} \right)}} + \dfrac{1}{{{2^2}\left( {2!} \right)}} + ......\infty {\text{ }} - - - \left( 5 \right)$
because it is an infinite series so $n \to \infty$
Now, we know that
${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....$
Here, $x = \dfrac{1}{2}$
$\therefore {e^{\dfrac{1}{2}}} = 1 + \dfrac{1}{{2\left( {1!} \right)}} + \dfrac{{{x^2}}}{{{2^2}\left( {2!} \right)}} + \dfrac{{{x^3}}}{{{2^3}\left( {3!} \right)}} + ....$
So, from equation $\left( 5 \right)$
$S = {e^{\dfrac{1}{2}}}$
$\Rightarrow S = \sqrt e$
Hence, option (3) is correct.

Note:
One may go wrong if he/she solves all the terms given in the series and then try to solve series. So, the given series is already following a particular order, observe it and hence solve it.
Also, there is an alternative way to solve this problem
The given series is
$S = 1 + \dfrac{1}{{2!}} + \dfrac{{1 \cdot 3}}{{4!}} + \dfrac{{1 \cdot 3 \cdot 5}}{{6!}} + \cdot \cdot \cdot {\text{ }} - - - - \left( 1 \right)$
$\Rightarrow S = 1 + \dfrac{1}{{2 \times 1}} + \dfrac{{1 \cdot 3}}{{4 \times 3 \times 2 \times 1}} + \dfrac{{1 \cdot 3 \cdot 5}}{{6 \times 5 \times 4 \times 3 \times 2 \times 1}} + \cdot \cdot \cdot \cdot$
On cancelling,
$\Rightarrow S = 1 + \dfrac{1}{{2 \times 1}} + \dfrac{{1 \cdot 3}}{{4 \times 2}} + \dfrac{{1 \cdot 3 \cdot 5}}{{6 \times 4 \times 2}} + \cdot \cdot \cdot \cdot$
$\Rightarrow S = 1 + \dfrac{1}{{2 \times 1}} + \dfrac{{1 \cdot 3}}{{{2^2} \times 2}} + \dfrac{{1 \cdot 3 \cdot 5}}{{6 \times {2^2} \times 2}} + \cdot \cdot \cdot \cdot$
On simplification,
$\Rightarrow S = 1 + \dfrac{1}{{2 \times 1!}} + \dfrac{{1 \cdot 3}}{{{2^2} \times 2!}} + \dfrac{{1 \cdot 3 \cdot 5}}{{{2^3} \times 3!}} + \cdot \cdot \cdot \cdot {\text{ }} - - - \left( 2 \right)$
Now, we know that
${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....$
Here, $x = \dfrac{1}{2}$
$\therefore {e^{\dfrac{1}{2}}} = 1 + \dfrac{1}{{2\left( {1!} \right)}} + \dfrac{{{x^2}}}{{{2^2}\left( {2!} \right)}} + \dfrac{{{x^3}}}{{{2^3}\left( {3!} \right)}} + ....$
So, from equation $\left( 2 \right)$
$S = {e^{\dfrac{1}{2}}}$
$\Rightarrow S = \sqrt e$
Hence, option (3) is correct.