A special sequence

$\text{claim}:a_n= k-\sum_{s=2}^n \dfrac{1}{s!} \forall n\geq 2$

proof(by induction) : the base case of $n=2$ holds true. $a_{n+1} = a_n-\dfrac{1}{(n+1)!} =k-\left(\sum_{s=2}^n \dfrac{1}{s!}\right)-\dfrac{1}{(n+1)!}= k-\sum_{s=2}^{n+1} \dfrac{1}{s!}$ and hence by induction this is proven.

for the series to converge, we need $\lim_{n\to \infty} a_n = 0 \to k= \lim_{n\to \infty} \sum_{s=2}^n \dfrac{1}{s!} = \boxed{ e-2}\to a_n =e-2-\sum_{s=2}^n \dfrac{1}{s!}$ we can make the sum nicer by using the taylor series of $e^x$ evaluated at $x=1$ . so $\sum_{m=1}^\infty \left(e-2-\sum_{s=2}^m \dfrac{1}{s!}\right) = \sum_{m=1}^\infty \sum_{s=m+1}^\infty \dfrac{1}{s!}= \sum_{s=2}^\infty \sum_{m=1}^{s-1} \dfrac{1}{s!}= \sum_{s=2}^\infty\dfrac{s-1}{s!} = \sum_{s=2}^\infty\left(\dfrac{1}{(s-1)!}-\dfrac{1}{s!}\right) = \boxed{1}$

$\ a_{n}=k-\sum_{i=2}^n \frac{1}{i!}$

As can be proven by induction. Therefore

$\lim_{n \rightarrow \infty} \ a_{n}=k+2-e$ Since $e=\sum_{i=0}^{\infty} \frac{1}{i!}$

The partial sum is by induction

$\ S_{n}=n(k-\sum_{i=2}^n \frac{1}{i!})+\sum_{i=2}^n \frac{i-1}{i!}=na_{n}+\sum_{i=2}^n \frac{i-1}{i!}$ .

For the sum to comverge first term of right end side must be zero. Hence $\ k=e-2$

In this way:

$\lim_{n \rightarrow \infty} na_{n}=\sum_{i=n}^{\infty} \frac{n}{i!}=0$

In order to evaluate the 2nd term use the identity

$e^x=\sum_{i=0}^{\infty} \frac{x^i}{i!}$

Divide the identity by $\ x$ and take the first derivative:

$\frac{e^x}{x}-\frac{e^x}{x^2}=-\frac{1}{x^2}+\sum_{i=2}^{\infty} \frac{(i-1)x^{i-2}}{i!}$

Using $\ x=1$ in the above expression yelds

$0=-1+\sum_{i=2}^{\infty} \frac{(i-1)}{i!}$

$1=\sum_{i=2}^{\infty} \frac{(i-1)}{i!}$

Therefore

$K= \lim_{n \rightarrow \infty} \ S_{n}=1$

$K+k =e-1=\boxed{1.718}$