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Find the sum of the infinite series

$$ 1 + \frac{1+2}{2!} + \frac{1+2+2^2}{3!} +\frac{1+2+2^2+2^3}{4!}+... ....$$

What I have done let

$$ S = \underbrace{\frac{1}{1!}}_{\text{1st Term}} + \underbrace{\frac{1+2}{2!}}_{\text{2nd Term}} + \underbrace{\frac{1+2+2^2}{3!}}_{\text{3rd Term}} + \underbrace{\frac{1+2+2^2+2^3}{4!}}_{\text{4th Term}} +... ....$$

I can see the denominator can be written as such but I'm not sure how to manipulate the numerator?

$$ S = \sum^{\infty}_{n=1} \frac{\text{?}}{n!} $$

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    The numerator is $2^n - 1$ – 2017-02-11

1 Answers 1

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The numerator is simply geometric progression, $$1 + 2 + 2^2 + \cdots + 2^{n-1} = 2^n - 1$$ Therefore, $$\sum_{n=1}^{\infty} \frac{2^n - 1}{n!} = \sum_{n = 1}^{\infty}\frac{2^n}{n!} - \sum_{n = 1}^{\infty}\frac{1}{n!} = (e^2 - 1) - (e - 1) = e^2 -e$$

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    How did you get the infinite sum of the first sum was $e^2-1$ and the second to be $e-1$ ? – 2017-02-11
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    Thatt is standard result you can find .$\sum_{n=0}^\infty \frac{x^n}{n!}=e^x$ – 2017-02-11