Binomial Theorem, Exponential and Logarithmic Series
1. Exponential Series
Some important notes:
\[{e^x} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ..........\infty \]
\[{e^{ - x}} = 1 - \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} - \frac{{{x^3}}}{{3!}} + ..........\infty \]
\[\frac{{{e^x} + {e^{ - x}}}}{{2!}} = \left[ {1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} + .........\infty } \right]\]
\[\frac{{{e^x} - {e^{ - x}}}}{{2!}} = \left[ {\frac{x}{{1!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} + .........\infty } \right]\]
\[e = \sum\limits_{n = 0}^\infty {\frac{1}{{n!}} = } \;1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + ..........\infty \]
\[{e^{ - 1}} = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{n!}} = \;} 1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + ..........\infty \]
\[\frac{{e + {e^{ - 1}}}}{{2!}} = \sum\limits_{n = 0}^\infty {\frac{1}{{[2n]!}}} \; = \;\left( {1 + \frac{1}{{2!}} + \frac{1}{{4!}} + .........\infty } \right)\]
\[\frac{{e - {e^{ - 1}}}}{{2!}} = \sum\limits_{n = 0}^\infty {\frac{1}{{(2n + 1)!}}} \; = \;\left( {\frac{1}{{1!}} + \frac{1}{{3!}} + \frac{1}{{5!}} + .........\infty } \right)\]
Example 01: Find the sum of the series \(\frac{2}{{1!}} + \frac{4}{{3!}} + \frac{6}{{5!}} + ......... + \infty \) VIEW SOLUTION