A Problem Based On One Of The Most Important Concept In Mathematics

image containing math problem statement
Do the terms of this series look familiar to you? This problem can expose you to a very important technique to solve series like this.
We will begin solving this problem starting with the definition of $e^x$, as given   :

$e^x = \dfrac{x^0}{0!} + \dfrac{x^1}{1!} + \dfrac{x^2}{2!} + ... \infty$

$\Rightarrow e^x = 1 + \dfrac{x^1}{1!} + \dfrac{x^2}{2!} + ... \infty$

$\Rightarrow e^x = 1 + x \left(\dfrac{x^0}{1!} + \dfrac{x^1}{2!} +  \dfrac{x^2}{3!} + ... \infty \right)$

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$\Rightarrow e^x = 1 + x \left(\dfrac{x^0}{1!} + \dfrac{x^1}{2!} + \dfrac{x^2}{3!} + ... \infty \right)$

Differentiating both sides w.r.t. $x$:
$\dfrac{d}{dx}(e^x) = 0 + \dfrac{d}{dx} \left[x \left(\dfrac{x^0}{1!} + \dfrac{x^1}{2!} + \dfrac{x^2}{3!} + ... \infty \right)\right]$

$\Rightarrow e^x = \left(\dfrac{x^0}{1!} + \dfrac{x^1}{2!} + \dfrac{x^2}{3!} + ... \infty \right) + x\left(0 + \dfrac{x^0}{2!} +  \dfrac{2x^1}{3!} +... \infty \right)$

$\Rightarrow e^x = \dfrac{1}{x}\left(\dfrac{x^1}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + ... \infty \right) + x\left(0 + \dfrac{x^0}{2!} + \dfrac{2x^1}{3!} +... \infty \right)$

$\Rightarrow e^x = \dfrac{1}{x} \left(1 + \dfrac{x^1}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!}  + ... \infty - 1\right) + x\left(\dfrac{x^0}{2!} + \dfrac{2x^1}{3!} + \dfrac{3x^2}{3!} +... \infty \right)$

$\Rightarrow e^x = \dfrac{1}{x} \left(e^x - 1\right) + x\left(\dfrac{x^0}{2!} + \dfrac{2x^1}{3!} + \dfrac{3x^2}{4!} +... \infty \right)$

Now substituting $x = 1$, we get:

$e^1 = \dfrac{1}{1} \left(e^1 - 1\right) + 1\left(\dfrac{1^0}{2!} + \dfrac{2 \cdot 1^1}{3!} + \dfrac{3 \cdot 1^2}{4!} +... \infty \right)$

$\Rightarrow \cancel{e^1} = \cancel{e^1} - 1 + \left(\dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} +... \infty \right)$

$\therefore \dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} +... \infty = 1$