The series \(\mathop \sum \limits_{n = 0}^\infty \frac{1}{{n!}}\)
A. 2 In 2
B. √2
C. 2
D. e
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Right Answer is: D
SOLUTION
Given:
Expanding the given expression, we can write:
\(\mathop \sum \limits_{n = 0}^\infty \frac{1}{{\angle n}} = 1 + \frac{1}{{\angle 1}} + \frac{1}{{\angle 2}} + \frac{1}{{\angle 3}} + \frac{1}{{\angle 4}} + \ldots \)
\(\mathop \sum \limits_{n = 0}^\infty \frac{1}{{\angle n}} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{{24}} + \ldots \) --- (1)
The series expansion of exponential function is given by:
\({e^x} = 1 + x + \frac{{{x^2}}}{{\angle 2}} + \frac{{{x^3}}}{{\angle 3}} + \frac{{{x^4}}}{{\angle 4}} + \ldots \)
\({e^x} = 1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + \frac{{{x^4}}}{{24}} + \ldots \)
Putting x = 1
\({e^1} = e = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{{24}} + \ldots \) ---(2)
From (1) and (2), we get:
\(e = \mathop \sum \limits_{n = 0}^\infty \frac{1}{{\angle n}}\)
Or
\(\mathop \sum \limits_{n = 0}^\infty \frac{1}{{\angle n}} = e\)
Hence option 4 is correct.