Summation of expression with factorials

Question

Is there an easy way to evaluate the following sum without summing all the terms one by one?

$$\Large\sum_{i=1}^{10} {i! \cdot i}$$

Answer 1

The following sum:

$$\large\sum_{i=1}^{n}{i! \cdot i}$$

can be proven to be equal to $(n+1)! - 1$ using inductive reasoning.

For some property, $$P(n) = \large\sum_{i=1}^{n}{i! \cdot i}$$ using $P(1) = (1+1)! - 1$ to be a basis, assume $P(k) = (k+1)! - 1$.

Then, prove $P(k) \implies P(k + 1)$, which means $P(k+1) = (k+2)! - 1$, to be true.

\begin{align*} \sum_{i = 1}^{k + 1} i! \cdot i & = \sum_{i = 1}^{k} i! \cdot i + (k + 1)(k + 1)!\\ & = (k + 1)! - 1 + (k + 1)(k + 1)! & \text{by the induction hypothesis}\\ & = (1 + k + 1)(k + 1)! - 1\\ & = (k + 2)(k + 1)! - 1\\ & = (k + 2)! - 1 \end{align*}

Therefore, $P(k) \implies P(k + 1)$. $P(n)$ holds $\forall n \in \mathbb{Z_+}$.

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We can now rewrite the original sum as

$$\large\sum_{i=1}^{n}{i! \cdot i} = (n + 1)! - 1$$

Therefore, $n=10$ is simply

$$\large\sum_{i=1}^{10}{i! \cdot i} = (10 + 1)! - 1 = 11! - 1 = 39916799$$