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Algebra Difficulty 4.6 AIME Prove it Belarus

Prove the inequality
11!+12!+13!+⋯+12022!>122!+223!+324!+⋯+202222023!. \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{2022!} > \frac{1^2}{2!} + \frac{2^2}{3!} + \frac{3^2}{4!} + \dots + \frac{2022^2}{2023!}.

Solution

Let us prove that ∑n=12022n2(n+1)!−∑n=120221n!<0\sum_{n=1}^{2022} \frac{n^2}{(n+1)!} - \sum_{n=1}^{2022} \frac{1}{n!} < 0, which is equivalent to the required. Note that
n2(n+1)!−1n!=n(n+1)−2(n+1)+1(n+1)!=1(n−1)!−2n!+1(n+1)! \frac{n^2}{(n+1)!} - \frac{1}{n!} = \frac{n(n+1) - 2(n+1) + 1}{(n+1)!} = \frac{1}{(n-1)!} - \frac{2}{n!} + \frac{1}{(n+1)!}
With this identity in mind, we make the following transformations:
∑n=12022n2(n+1)!−∑n=120221n!=∑n=020211n!−2∑n=120221n!+∑n=220231n!==10!−11!−12022!+12023!=12023!−12022!<0. \begin{aligned} & \sum_{n=1}^{2022} \frac{n^2}{(n+1)!} - \sum_{n=1}^{2022} \frac{1}{n!} = \sum_{n=0}^{2021} \frac{1}{n!} - 2 \sum_{n=1}^{2022} \frac{1}{n!} + \sum_{n=2}^{2023} \frac{1}{n!} = \\ & = \frac{1}{0!} - \frac{1}{1!} - \frac{1}{2022!} + \frac{1}{2023!} = \frac{1}{2023!} - \frac{1}{2022!} < 0. \end{aligned}

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