Question

Show that $1 + \dfrac{1}{{2!}} + \dfrac{1}{{4!}} + ........ = \dfrac{1}{2}\left( {e + \dfrac{1}{e}} \right)$

Answer 1

Hint: To prove the expression given in the question, we will use the expression of binomial expansion of ${e^x}$. Then we will substitute the value of $x$ as 1. Next, we will use the expression of binomial expansion of ${e^{ - x}}$ and again we will substitute the value of $x$ as 1. Now, on adding both expressions obtained from the binomial expansion of ${e^x}$ and ${e^{ - x}}$ , we will prove the above result.


Complete step-by-step answer:
We will use the expression of binomial expansion of ${e^x}$ which can be written as:
\[{e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + .............\]
 We will also use the expression of binomial expansion of ${e^{ - x}}$ which can be written as
 \[{e^{ - x}} = 1 - x + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + .............\]
Step by step answer:
We know that binomial expansion of the function ${e^x}$ can be expressed as
\[{e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + .............\]
 We will substitute 1 for $x$ in the above expression.
\[e = 2 + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + \dfrac{1}{{4!}} + .......\] ……(i)
 We also know that the binomial expansion of the function ${e^{ - x}}$ can be expressed as
\[{e^{ - x}} = 1 - x + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + .............\]
 We will substitute 1 for $x$ in the above expression.
\[{e^{ - 1}} = \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}} + .......\] ……(ii)
 Now we will add equation (i) and (ii), we will get
\[\begin{array}{l}
e + {e^{ - 1}} = 2 + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + \dfrac{1}{{4!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}} + ......\\
e + {e^{ - 1}} = 2 + \dfrac{1}{{2!}} + \dfrac{1}{{4!}} + ......
\end{array}\]
 Next, we will divide both side of the above expression by 2. Then the above expression can be it expressed as:
\[\begin{array}{l}
\dfrac{1}{2}\left( {e + \dfrac{1}{e}} \right) = \dfrac{1}{2}\left( {2 + 2\left( {\dfrac{1}{{2!}}} \right) + 2\left( {\dfrac{1}{{4!}}} \right) + ......} \right)\\
\dfrac{1}{2}\left( {e + \dfrac{1}{e}} \right) = 1 + \dfrac{1}{{2!}} + \dfrac{1}{{4!}} + ........
\end{array}\]
Hence proved.

Note: To prove the expression given in the question, we should have prior knowledge of the binomial expansion of exponential functions. In this question we are using the right-hand expression to prove the result. We will have to take care of signs of variables before writing the expressions.