Approximation of $\sum_{k=1}^\infty \frac{1}{T_k}$ where $T_k$ are the tribonacci numbers So I have heard about the reciprocal Fibonacci constant, which is the sum of the reciprocals of the Fibonacci numbers. A natural question is to ask what we get when we add the reciprocals of the Tribonacci numbers $T_k$ (which satisfy the recurrence relation $T_k=T_{k-1}+T_{k-2}+T_{k-3}$ and $T_0=0$ and $T_1=T_2=1$). I tried to put $\sum_{k=1}^\infty \frac{1}{T_k}$ into Wolfram Alpha but it says that this diverges. Of course that's just absurd since the Tribonacci numbers grow quicker than the Fibonacci numbers. My question is, may you help give me an approximation of this constant up to $3$ decimal digits? I don't have any advanced math software like Mathematica. And is my claim that the sum converges true? My only reason why this converges is the fact that it seems that the Tribonacci numbers grow quicker than the Fibonacci numbers, which seems obvious but I don't know for sure.
Thanks for reading the whole post!
 A: Assuming the sum does converge, we can write up a little Python script to approximate the value (Thanks to Calvin Khor and G Cab for the improvements):
a, b, c = 0, 1, 1
s = 2

for _ in range(1000):
    a, b, c = b, c, a + b + c
    s += 1 / c

print(s)

We see that the sum of the first $1002$ terms is
$$
\sum_{k=1}^{\infty}\frac{1}{T_k} \approx\sum_{k=1}^{1002}\frac{1}{T_k} \approx 3.061229085105492
$$
A: It can be shown that
$$ \sum_{k=n}^{\infty} \frac{1}{T_{k}} < \frac{1}{T_{n} - T_{n-1} - 1}, $$
which, for $n>4$, works well. Now, for $n>0$, the sum takes the form
$$ \sum_{k=1}^{\infty} \frac{1}{T_{k}} < \frac{1}{T_{1}} + \frac{1}{T_{2}} + \frac{1}{T_{3}} + \frac{1}{T_{4}} + \frac{1}{T_{5} - T_{4} - 1} = 3.25. $$
This can be partially found in the article by Anantakitpaisal and Kuhapatanakul.
Numerically the value of the sum is
$$3.0612290851054928774253826071147789249412794648974344700718107630658978210...$$
which, for this case, was obtained from Sagemath by use of the code
@CachedFunction
def T(n): # Tribonacci numbers
    if (n<3): return (0,1,1)[n]
    else: return T(n-1) + T(n-2) + T(n-3)
@CachedFunction
def S(n): return 1/T(1) if (n==1) else 1/T(n) + S(n-1)
numerical_approx(S(400), digits=100)

which can be run in SageCell at SageCell. Note that on the Sagecell site, by changing the language setting to Python, the code in another answer can also be run to obtain the numerical value by use of Python.
