Question

The exponential function \[{e^x}\] can be defined as a power series as:
\[{e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ...\]
Can you use this definition to evaluate \[\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}{e^{ - 0.2}}}}{{n!}}} \] ?

Answer 1

Hint: In the above question, we are given the exponential function \[{e^x}\] and its expansion in the form series. We have to use the power series to evaluate another series \[\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}{e^{ - 0.2}}}}{{n!}}} \] . Here, we can separate the function in two parts as \[{e^{ - 0.2}}\] and \[\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}} \] so that we can easily evaluate \[\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}} \] using the given power series.

Complete step by step answer:
Given the power series of the exponential function \[{e^x}\] is,
\[ \Rightarrow {e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ...\]
We have to use the above power series to evaluate the function,
\[ \Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}{e^{ - 0.2}}}}{{n!}}} \]
We can also write the above function as,
\[ \Rightarrow {e^{ - 0.2}}\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}} \]
This is the summation over $n$. So, taking out $e^{-0.2}$ out of the summation won’t change the value.
Now, consider \[\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}} \] .
Now since,
\[{e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ...\]
Using the above given expansion of \[{e^x}\] in the expression \[\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}} \] , we can write,
\[ \Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}} = {e^{0.2}}\]
Now, putting \[\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}} = {e^{0.2}}\] in the expression \[{e^{ - 0.2}}\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}} \] we get,
\[ \Rightarrow {e^{ - 0.2}} \cdot {e^{0.2}}\]
That gives,
\[ \Rightarrow \dfrac{{{e^{0.2}}}}{{{e^{0.2}}}}\]
Hence,
\[ \Rightarrow 1\]
That is the required value of the above given function \[\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}{e^{ - 0.2}}}}{{n!}}} \] .
Therefore, the value of the series \[\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}{e^{ - 0.2}}}}{{n!}}} \] is \[1\].

Note:
The number \[e\] given in the above question is known as the Euler’s number. The function of the Euler’s number of the type \[{e^x}\] with the independent variable \[x\] ranging over the entire real number line as the exponent of the positive constant number \[e\] . It is known as a natural exponential function or just as an exponential function.
The Euler’s number \[e\] is a constant having a value of nearly \[e = 2.718281828...\]
The inverse function of the natural exponential function \[{e^x}\] is \[{\log _e}x\] .
This is known as the natural logarithmic function or just as the logarithmic function, also denoted by \[\ln x\] .