Question

Solve $\sum\limits_{r = 1}^\infty {\dfrac{{1 + a + {a^2} + {a^3} + .... + {a^{r - 1}}}}{{r!}}} $

Answer 1

Hint: First we will convert the GP into a numerator in the form of a sum of terms of a GP. Hence we will get a simplified form of the sum of the series. Then, on simplifying the given equation by performing required operations. Then, we have to convert the sum of the series into such a form that we can deduce it into a shorter form of the series.

Complete step-by-step answer:
Given, $\sum\limits_{r = 1}^\infty {\dfrac{{1 + a + {a^2} + {a^3} + .... + {a^{r - 1}}}}{{r!}}} $
In the numerator we can clearly observe it to be a GP with first term as $1$ and common difference $a$, extending till $r$ terms.
So, we know, the formula for the sum of terms of a GP is $\dfrac{{p({q^n} - 1)}}{{(q - 1)}}$.
Therefore, using this formula in the above equation, we get,
$ \Rightarrow \sum\limits_{r = 1}^\infty {\dfrac{{1 + a + {a^2} + {a^3} + .... + {a^{r - 1}}}}{{r!}}} = \sum\limits_{r = 1}^\infty {\dfrac{{\dfrac{{1\left( {{a^r} - 1} \right)}}{{\left( {a - 1} \right)}}}}{{r!}}} $
$ \Rightarrow \sum\limits_{r = 1}^\infty {\dfrac{{1 + a + {a^2} + {a^3} + .... + {a^{r - 1}}}}{{r!}}} = \sum\limits_{r = 1}^\infty {\dfrac{{{a^r} - 1}}{{r!\left( {a - 1} \right)}}} $
We can write the above equation in the form,
$ \Rightarrow \sum\limits_{r = 1}^\infty {\dfrac{{1 + a + {a^2} + {a^3} + .... + {a^{r - 1}}}}{{r!}}} = \dfrac{1}{{\left( {a - 1} \right)}}\sum\limits_{r = 1}^\infty {\dfrac{{{a^r} - 1}}{{r!}}} $
Now, dividing the numerator by the denominator, we get,
$ \Rightarrow \sum\limits_{r = 1}^\infty {\dfrac{{1 + a + {a^2} + {a^3} + .... + {a^{r - 1}}}}{{r!}}} = \dfrac{1}{{\left( {a - 1} \right)}}\sum\limits_{r = 1}^\infty {\left[ {\dfrac{{{a^r}}}{{r!}} - \dfrac{1}{{r!}}} \right]} $
Now, expanding the terms, we get,
$ \Rightarrow \sum\limits_{r = 1}^\infty {\dfrac{{1 + a + {a^2} + {a^3} + .... + {a^{r - 1}}}}{{r!}}} = \dfrac{1}{{\left( {a - 1} \right)}}\left[ {\left\{ {\dfrac{a}{{1!}} + \dfrac{{{a^2}}}{{2!}} + \dfrac{{{a^3}}}{{3!}} + .....} \right\} - \left\{ {\dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ....} \right\}} \right]$
Now, we know, the expansion of ${e^a} = 1 + \dfrac{a}{{1!}} + \dfrac{{{a^2}}}{{2!}} + \dfrac{{{a^3}}}{{3!}} + .....$
$ \Rightarrow {e^a} - 1 = \dfrac{a}{{1!}} + \dfrac{{{a^2}}}{{2!}} + \dfrac{{{a^3}}}{{3!}} + .....$
And, the expansion, $e = 1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + .....$
$ \Rightarrow e - 1 = \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + .....$
Therefore, substituting the expansions in the equation, we get,
$ \Rightarrow \sum\limits_{r = 1}^\infty {\dfrac{{1 + a + {a^2} + {a^3} + .... + {a^{r - 1}}}}{{r!}}} = \dfrac{1}{{\left( {a - 1} \right)}}\left[ {\left\{ {{e^a} - 1} \right\} - \left\{ {e - 1} \right\}} \right]$
Opening the brackets and simplifying, we get,
$ \Rightarrow \sum\limits_{r = 1}^\infty {\dfrac{{1 + a + {a^2} + {a^3} + .... + {a^{r - 1}}}}{{r!}}} = \dfrac{1}{{\left( {a - 1} \right)}}\left[ {{e^a} - 1 - e + 1} \right]$
$ \Rightarrow \sum\limits_{r = 1}^\infty {\dfrac{{1 + a + {a^2} + {a^3} + .... + {a^{r - 1}}}}{{r!}}} = \dfrac{1}{{\left( {a - 1} \right)}}\left[ {{e^a} - e} \right]$
$ \Rightarrow \sum\limits_{r = 1}^\infty {\dfrac{{1 + a + {a^2} + {a^3} + .... + {a^{r - 1}}}}{{r!}}} = \dfrac{{{e^a} - e}}{{a - 1}}$
Therefore, the solution of the sum of the series is $\dfrac{{{e^a} - e}}{{a - 1}}$.
So, the correct answer is “Option B”.

Note: The major part where we make mistakes is in detecting the short expansions of the series that we are supposed to substitute. We must keep in account some of the commonly used series expansions or either try to convert the series in the form of GP or AP in order to easily substitute in the form of sum of terms of AP or GP in order to shorten the series to operate easily.