$\frac{1}{1!} + \frac{4}{2!} + \frac{7}{3!} + \frac{10}{4!} + \dots \infty = $

  • A
    $e + 4$
  • B
    $2 + e$
  • C
    $3 + e$
  • D
    $e$

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The value of $\sum_{r=2}^{\infty} \frac{1+2+\dots+(r-1)}{r !}$ is:

The value of $1 - \log 2 + \frac{(\log 2)^2}{2!} - \frac{(\log 2)^3}{3!} + \dots$ is

The sum of the series $\frac{1}{2 !} + \frac{1+2}{3 !} + \frac{1+2+3}{4 !} + \ldots$ is equal to :

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

Difficult
View Solution

$\left( {1 + \frac{1}{{2!}} + \frac{1}{{4!}} + \dots} \right) \left( {1 + \frac{1}{{3!}} + \frac{1}{{5!}} + \dots} \right) = $

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