# Given $S_n = \sum \dots$ and $a_n = \sum \dots$ prove that $a_n = S_n + {1\over n\cdot n!}$

I'm trying to solve the following problem:

Let: \begin{align} S_n &= 2 + {1\over2!} + {1\over3!} + {1\over 4!} + \dots + {1\over n!} \\ a_n &= 3 - {1\over 1\cdot2\cdot2!} - {1\over 2\cdot3\cdot3!} - \dots - {1\over (n-1)\cdot n\cdot n!} \end{align} Prove that: $$a_n = S_n + {1\over n\cdot n!}$$

My thought on this are:

Manipulating the sums here would become pretty hard so I decided to use another approach. I've tried to look what the first terms of each sum are going to be in order to observe a pattern:

\begin{align} S_1 &= 2 \\ S_2 &= 2 + {1\over 2!} = S_1 + {1\over 2!} \\ S_3 &= 2 + {1\over 2!} + {1\over 3!} = S_2 + {1\over 3!}\\ &\vdots \\ S_{n+1} &= S_n + {1\over (n+1)!} \end{align}

A similar thing is done to $$a_n$$:

$$a_{n+1} = a_n - {1\over n\cdot (n+1) \cdot (n+1)!}$$

So potentially if we could find closed forms of both $$S_n$$ and $$a_n$$ it would become easier to reason about them.

By the way, I've observed both of the sums approach $$e$$ at infinity but from different sides. $$S_n$$ is starting from $$2$$ adding more terms as $$n$$ grows, while $$a_n$$ starts from 3 and decreases the sum as $$n$$ grows. So that will be our initial conditions for both recurrences:

$$S_1 = 2 \\ a_1 = 3$$

The problem just reduced to finding closed forms of the recurrences. Both of them are non-homogenous and here is where I got stuck. Solving for homogenous part is easy since roots of both characteristic equations $$\lambda_{a_n} = \lambda_{S_n} = 1$$.

I couldn't find a particular solution for them. I've tried yet another approach expressing $$S_{n+1}$$ and $$S_{n+2}$$ trying to get rid of the $$1\over (n+1)!$$ or at least change its form so the guess for particular solution is more obvious.

My questions are:

1. Is it even a valid approach to solve the problem?
2. If so then how could I find a closed form for the recurrences? Especially i'm interested in finding a particular solution. (Yet a complete flow is very appreciated and encourages learning by example)
3. Not sure how to phrase that better, but does there exist a table of "guesses" for particular solution of the recurrences in some form of $$x_{n+1} = x_n + F(n)$$. Like if the "free term" is a constant say $$F(n) = 2$$ then the guess for particular solution is some constant $$B$$.

Please excuse me if there is some vagueness or inaccuracy in the terminology, English is not my native language and I have almost no background in Maths.

• It seems unlikely to find a simple closed form, because, as you noticed, the limit converges to $e$. – user600464 Oct 5 '18 at 18:59
• @user600464 is that because $e$ is irrational and that is why no "simple" formula for an irrational number may be obtained? – roman Oct 5 '18 at 19:02
• There is actually a simple formula, the limit $\frac{n^n}{(n- 1)^n}$. However, to compare to this sums, we would need many terms – user600464 Oct 5 '18 at 19:34

$$a_2 - S_2 = 3- 1/4 - (1 + 1/1! + 1/2!)$$ [I think there is some typo in your expression of $$S_n$$] $$= 1/4 = 1/(2 \cdot 2!)]$$. Suppose the equation holds for $$n$$. Then for $$n+1$$, \begin{align*} a_{n+1} - S_{n+1} &= a_n - S_n - \frac 1 {n(n+1) \cdot (n+1)!} - \frac 1{(n+1)!} \\ &= \frac 1 {n!n }-\frac 1 {n(n+1) \cdot (n+1)!} - \frac 1{(n+1)!} \\ &= \frac 1 {(n+1)! (n+1)} \left( \frac {(n+1)^2}n -\frac 1n -(n+1) \right)\\ &= \frac 1 {(n+1)! (n+1)}. \end{align*} Thus the equation holds for all $$n \geqslant 2$$ by induction principle.
\begin{align} \color{red}{a_{n}-S_{n}} &=\left(3-\sum_{k=2}^{n}\frac{1}{(n-1)\cdot n\cdot n!}\right)-\left(2+\sum_{k=2}^{n}\frac{1}{n!}\right) \\[2mm] &=1-\sum_{k=2}^{n}\frac{1}{n!}\left(1+\frac{1}{(n-1)\cdot n}\right) =1-\sum_{k=2}^{n}\frac{1}{n!}\left(1+\frac{1}{n-1}-\frac{1}{n}\right) \\[2mm] &=1-\sum_{k=2}^{n}\frac{1}{n!}\left(\frac{n}{n-1}-\frac{1}{n}\right) =1-\sum_{k=2}^{n}\left(\frac{1}{(n-1)\cdot (n-1)!}-\frac{1}{n\cdot n!}\right) \\[2mm] &=1-\frac{1}{1\cdot1!}+\frac{1}{2\cdot2!}-\frac{1}{2\cdot2!}+\frac{1}{3\cdot3!}-\cdots+\frac{1}{n\cdot n!}=\color{red}{\frac{1}{n\cdot n!}}\quad\left\{\text{telescoping}\right\} \end{align}