Solving for $g(k)$ in the implicit equation $(k+1)g(k)+k!=g(k+1)$

Question

Is there any good solution method to find the function $g(k)$ where $k \in \mathbb{N}$ that solves the following implicit equation? $$(k+1)g(k)+k!=g(k+1)$$ I found that using $g(k) = k! - (k-1)!$ evaluates the left side as $$(k+1)(k!-(k-1)!)+k!$$ $$(k+1)!-(k+1)(k-1)!+k!$$ $$(k+1)!-k!-(k-1)!+k!$$ $$(k+1)!-(k-1)!$$ and the right side as $$(k+1)!-k!$$ which is close (depending on your definition of what close is). Is there any way to solve for $g(k)$ without just plugging in random functions?

Answer 1

Dividing throughout by $(k+1)!$, we obtain $$\frac{g(k+1)}{(k+1)!}-\frac{g(k)}{k!}=\frac{1}{k+1}.$$ Notice that the left hand side telescopes. In particular, summing the equalities $$\begin{split} \frac{g(2)}{2!}-\frac{g(1)}{1!}&=\frac{1}{2}\\ \frac{g(3)}{3!}-\frac{g(2)}{2!}&=\frac{1}{3}\\ \frac{g(4)}{4!}-\frac{g(3)}{3!}&=\frac{1}{4}\\ &\vdots\\ \frac{g(k)}{k!}-\frac{g(k-1)}{(k-1)!}&=\frac{1}{k} \end{split}$$ we obtain $$\frac{g(k)}{k!}=g(1)+\left(\frac12+\frac13+\dots+\frac1k\right).$$

Answer 2

HINT

divide by $(k+1)!$ and sum over $k$ we get $\forall k \in \mathbb{N}$: $g(k+1)/(k+1)! - g(k)/k!=1/k+1 $ Note $G(k)=g(k)/k!$

$\forall n \in \mathbb{N}\sum_{1\leq k \leq n} G(k)-G(k-1) = \sum_{1\leq k \leq n} 1/k $ therefore $G(n+1)-G(0)=\sum_{1\leq k \leq n} 1/k $ which means $g(n)=\sum_{1\leq k \leq n-1} n!/k$