Analytic continuation of the sum of the reciprocals of the $n$-bonacci sequences.

Question

In a previous question, I asked for an approximation of the sum of the reciprocals of the Tribonacci numbers. So I was wondering if there is a function for the sum of the reciprocals of the $n$-bonacci numbers. This would look like this: $$\varkappa(n)=\sum_{k=1}^\infty\frac{1}{N_{n_k}}$$ Where $N_n$ gives the $n$-bonacci sequence ($n=1$ gives $1, 1, 1, ...$, $n=2$ gives the fibonacci sequence, $n=3$ gives the tribonacci sequence, and so on). I graphed the first four points in this desmos graph. So what function could connect these points? And additionally, does $\varkappa$ have a lower bound? What is it? (I suspect it is $1.5$ or something near but I couldn't prove this.

Answer 1

The generating function of the standard $n$-bonacci numbers being $$f_n(x)=\frac{ x^{n-1}}{1-\sum_{k=1}^{n}x^k}=\frac{(1-x)\, x^{n-1}}{x^{n+1}-2 x+1}$$

the sum of their reciprocals $S_n$ write $$S_n=3+\epsilon_n$$

$$\left( \begin{array}{ccc} n & \epsilon_n & \log(\epsilon_n) \\ 2 & 0.35988566624317755317 & -1.0220 \\ 3 & 0.06122908510549287743 & -2.7931 \\ 4 & 0.01325565493979152335 & -4.3233 \\ 5 & 0.00310889227803160293 & -5.7735 \\ 6 & 0.00075409134353350541 & -7.1900 \\ 7 & 0.00018577050528064920 & -8.5910 \\ 8 & 0.00004610682714077771 & -9.9845 \\ 9 & 0.00001148523413006498 & -11.374 \\ 10 & 0.00000286615546739387 & -12.763 \\ 11 & 0.00000071589666140425 & -14.150 \\ 12 & 0.00000017889400983734 & -15.536 \\ \end{array} \right)$$

A quick and dirty linear regression based on the data for $3 \leq n \leq 25$ $$\log(\epsilon_n)=a-b\,n$$ gives, with $R^2=0.999994$,

$$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & 1.19140 & 0.02474 & \{1.13996,1.24284\} \\ b & 1.39135 & 0.00160 & \{1.38803,1.39467\} \\ \end{array}$$