The sum of the series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$ up to infinity is

  • A
    $e^{-1/2}$
  • B
    $e^{1/2}$
  • C
    $e^{-2}$
  • D
    $e^{-1}$

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Find the sum of the series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty$.

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For every real number $x$, let $f(x) = \frac{x}{1!} + \frac{3}{2!} x^2 + \frac{7}{3!} x^3 + \frac{15}{4!} x^4 + \dots$. Then the equation $f(x) = 0$ has

In the expansion of $(e^x - 1)(e^{-x} + 1)$,the coefficient of $x^3$ is

$\frac{1 \cdot 2}{1!} + \frac{2 \cdot 3}{2!} + \frac{3 \cdot 4}{3!} + \frac{4 \cdot 5}{4!} + \dots \infty = $

The sum of the infinite series $1+\frac{1}{2!}+\frac{1 \cdot 3}{4!}+\frac{1 \cdot 3 \cdot 5}{6!}+\dots$ is

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