Question
If $$x \ne 0,$$ then the sum of the series $$1 + \frac{x}{{2!}} + \frac{{2{x^2}}}{{3!}} + \frac{{3{x^3}}}{{4!}} + .....\,\infty $$ is
A.
$$\frac{{{e^x} + 1}}{x}$$
B.
$$\frac{{{e^x}\left( {x - 1} \right)}}{x}$$
C.
$$\frac{{{e^x}\left( {x - 1} \right) + 1}}{x}$$
D.
$$\frac{{{e^x}\left( {x - 1} \right) + 1 + x}}{x}$$
Answer :
$$\frac{{{e^x}\left( {x - 1} \right) + 1 + x}}{x}$$
Solution :
The general term of the series
$$\eqalign{
& \frac{x}{{2!}} + \frac{{2{x^2}}}{{3!}} + \frac{{3{x^3}}}{{4!}} + .....\,\infty {\text{ is}} \cr
& {T_n} = \frac{{n{x^n}}}{{\left( {n + 1} \right)!}},n = 1,2,.....,\infty \cr
& = \frac{{n + 1 - 1}}{{\left( {n + 1} \right)!}}{x^n} = \frac{{{x^n}}}{{n!}} - \frac{1}{x}\frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}} \cr
& \therefore 1 + \frac{x}{{2!}} + \frac{{2{x^2}}}{{3!}} + \frac{{3{x^3}}}{{4!}} + .....\,\infty \cr
& = 1 + \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n!}} - \frac{1}{x}\sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}}} } \cr
& = 1 + \left( {{e^x} - 1} \right) - \frac{1}{x}\left( {{e^x} - 1 - x} \right) \cr
& = \frac{{x{e^x} - {e^x} + 1 + x}}{x} = \frac{{\left( {x - 1} \right){e^x} + \left( {1 + x} \right)}}{x} \cr} $$