Question

The value of $\dfrac{\dfrac{1}{2!}+\dfrac{1}{4!}+\dfrac{1}{6!}+......\infty }{1+\dfrac{1}{3!}+\dfrac{1}{5!}+\dfrac{1}{7!}+......\infty }$ is equal to?
(a) $\dfrac{e+1}{e-1}$
(b) $\dfrac{e-1}{e+1}$
(c) $\dfrac{{{e}^{2}}-1}{{{e}^{2}}+1}$
(d) $\dfrac{{{e}^{2}}+1}{{{e}^{2}}-1}$

Answer 1

Hint: Assume the given expression as E. First of all write the expansion formula of ${{e}^{x}}$ given by the Maclaurin series as ${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{5}}}{5!}+....$. Substitute the value of x equal to 1 and assume the expression obtained as (i). Now, substitute x equal to -1 and assume the expression obtained as (ii). Add the two equations and divide both the sides with 2 to find the value of the numerator of the expression E. Similarly, subtract the two equations to find the value of the denominator of the expression E. Use the algebraic identities ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$ and $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ to get the answer.

Complete step by step solution:
Here we have been provided with the expression $\dfrac{\dfrac{1}{2!}+\dfrac{1}{4!}+\dfrac{1}{6!}+......\infty }{1+\dfrac{1}{3!}+\dfrac{1}{5!}+\dfrac{1}{7!}+......\infty }$ and we are asked to find its simplified value. Here we will use the expansion formula of ${{e}^{x}}$ to get the answer. Assuming the given expression as E we get,
$\Rightarrow E=\dfrac{\dfrac{1}{2!}+\dfrac{1}{4!}+\dfrac{1}{6!}+......\infty }{1+\dfrac{1}{3!}+\dfrac{1}{5!}+\dfrac{1}{7!}+......\infty }$
Now, we know that the expansion formula of the exponential function ${{e}^{x}}$ is given using the Maclaurin series as ${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{5}}}{5!}+....$. Substituting x = 1 we get,
$\begin{align}
  & \Rightarrow {{e}^{1}}=1+1+\dfrac{{{1}^{2}}}{2!}+\dfrac{{{1}^{3}}}{3!}+\dfrac{{{1}^{4}}}{4!}+\dfrac{{{1}^{5}}}{5!}+....\infty \\
 & \Rightarrow {{e}^{1}}=1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\dfrac{1}{5!}+....\infty .......\left( i \right) \\
\end{align}$
Substituting x = -1 we get,
$\begin{align}
  & \Rightarrow {{e}^{-1}}=1+\left( -1 \right)+\dfrac{{{\left( -1 \right)}^{2}}}{2!}+\dfrac{{{\left( -1 \right)}^{3}}}{3!}+\dfrac{{{\left( -1 \right)}^{4}}}{4!}+\dfrac{{{\left( -1 \right)}^{5}}}{5!}+....\infty \\
 & \Rightarrow {{e}^{-1}}=1-1+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}-\dfrac{1}{5!}+....\infty .......\left( ii \right) \\
\end{align}$
Adding equations (i) and (ii) and simplifying by cancelling the common terms we get,
$\begin{align}
  & \Rightarrow {{e}^{1}}+{{e}^{-1}}=2\left( 1+\dfrac{1}{2!}+\dfrac{1}{4!}+\dfrac{1}{6!}....\infty \right) \\
 & \Rightarrow \dfrac{{{e}^{1}}+{{e}^{-1}}}{2}=\left( 1+\dfrac{1}{2!}+\dfrac{1}{4!}+\dfrac{1}{6!}....\infty \right) \\
 & \Rightarrow \dfrac{{{e}^{1}}+{{e}^{-1}}}{2}-1=\left( \dfrac{1}{2!}+\dfrac{1}{4!}+\dfrac{1}{6!}....\infty \right)...........\left( iii \right) \\
\end{align}$
Subtracting equation (ii) from equation (i) and simplifying by cancelling the common terms we get,
$\begin{align}
  & \Rightarrow {{e}^{1}}-{{e}^{-1}}=2\left( 1+\dfrac{1}{3!}+\dfrac{1}{5!}+\dfrac{1}{7!}....\infty \right) \\
 & \Rightarrow \dfrac{{{e}^{1}}-{{e}^{-1}}}{2}=\left( 1+\dfrac{1}{3!}+\dfrac{1}{5!}+\dfrac{1}{7!}....\infty \right)...........\left( iv \right) \\
\end{align}$
Now, substituting the values of the numerator and the denominator obtained in equations (iii) and (iv) we get the expression E as: -
$\Rightarrow E=\dfrac{\dfrac{{{e}^{1}}+{{e}^{-1}}}{2}-1}{\dfrac{{{e}^{1}}-{{e}^{-1}}}{2}}$
Using the conversion ${{e}^{-1}}=\dfrac{1}{e}$ we get,
$\begin{align}
  & \Rightarrow E=\dfrac{\dfrac{e+\dfrac{1}{e}-2}{2}}{\dfrac{e-\dfrac{1}{e}}{2}} \\
 & \Rightarrow E=\dfrac{{{e}^{2}}+1-2e}{{{e}^{2}}-1} \\
\end{align}$
Using the algebraic identities ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$ and $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$, cancelling the common factors to simplify we get,
$\begin{align}
  & \Rightarrow E=\dfrac{{{\left( e-1 \right)}^{2}}}{\left( e-1 \right)\left( e+1 \right)} \\
 & \therefore E=\dfrac{\left( e-1 \right)}{\left( e+1 \right)} \\
\end{align}$

So, the correct answer is “Option b”.

Note: Note that it is impossible to calculate the value of the expression with the help of the expansion series of the exponential function. This is because we cannot keep on calculating the sum and the factorials up to infinite terms. There are some functions like $\sin x,\cos x,\tan x,\ln \left( 1+x \right)$ etc. whose expansion formula you must remember because they are further used in the topic ‘limits’.