Here is my proof. It's a bit longer and uglier than what I wanted. I assumed direct calculation of a few numbers (albeit not running a for loop over the entire series) was allowed. I suspect there is a much shorter approach.
FWIW, it seemed to me that a better posed problem would be to prove:
##\sum_{k=1}^{n} \frac{1}{a_k a_{k+2}} \lt 4## for arbitrary natural number n, or perhaps just prove ##\sum_{k=1}^{\infty} \frac{1}{a_k a_{k+2}} \lt 4##
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The inspiration for this post is:
##\sum_{k=2}^n \big(\frac{1}{k}\big)^2 \lt \sum_{k=2}^n \frac{1}{k(k-1)} = 1 - \frac{1}{n} \leq 1##
for natural numbers where ##2\leq k \lt n##
additionally, the fact that all numbers in our series are strictly positive allows us to make use of things like ##GM \leq AM##
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It is sometimes useful to calculate a few values out first. Values from ##a_3, a_4, a_5, a_6## and ##a_7## are shown directly in the end piece of this post.
Our goal is to prove
##\sum_{k=1}^{98} \frac{1}{a_k a_{k+2}} = \big(\sum_{k=1}^{4} \frac{1}{a_k a_{k+2}}\big) + \sum_{k=5}^{98} \frac{1}{a_k a_{k+2}} \lt 4##
via above mentioned calculations, we have:
##\big(\sum_{k=1}^{4} \frac{1}{a_k a_{k+2}}\big)= \frac{1}{a_1 a_{3}} + \frac{1}{a_2 a_{4}} + \frac{1}{a_3 a_{5}} + \frac{1}{a_4 a_{6}} \approx 2.187 \lt 2.2 ##
The proof is in some sense only 3 lines, with some justification provided at the end
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first:
##\sum_{k=1}^{98} \frac{1}{a_k a_{k+2}} \lt 2.2 + \sum_{k=5}^{98} \frac{1}{a_k a_{k+2}} \lt 2.2 + \frac{1}{2}\Big(\big(\sum_{k=5}^{98} \frac{1}{a_k^2} \big) + \big(\sum_{k=5}^{98} \frac{1}{ a_{k+2}^2}\big)\Big) \lt 2.2 + \frac{1}{2}\Big(2\big(\sum_{k=5}^{100} \frac{1}{a_{k}^2} \big)\Big) = 2.2 + \Big(\sum_{k=5}^{100} \big(\frac{1}{a_{k}}\big)^2\Big)##
by application of Cauchy Schwarz in additive form (or equivalently: ##GM \leq AM##).
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second:
if we set: ##r:= k-3## we can re-write the RHS as
##2.2 + \Big(\sum_{k=5}^{100} \big(\frac{1}{a_{k}}\big)^2\Big) = 2.2 + \Big(\sum_{r=2}^{97} \big(\frac{1}{a_{r+3}}\big)^2\Big)##
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finally:
##\sum_{k=1}^{98} \frac{1}{a_k a_{k+2}} \lt 2.2 + \Big(\sum_{r=2}^{97} \big(\frac{1}{a_{r+3}}\big)^2\Big) \leq 2.2+ \Big(\sum_{r=2}^{97} \frac{1}{r(r-1)}\Big) \leq 2.2 + \big(1\big) = 3.2 \lt 4##
which completes the proof.
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note: this readily generalizes to an infinite series where:
##\sum_{k=1}^{\infty} \frac{1}{a_k a_{k+2}} \lt 2.2 + \Big(\sum_{k=2}^{\infty} \frac{1}{k(k-1)}\Big) = 3.2 \lt 4##
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supporting info:
proof that ##\sum_{r=2}^{n}\big(\frac{1}{a_{r+3}}\big)^2 \leq \sum_{r=2}^{n}\frac{1}{r(r-1)} ##
for any natural number ##n \geq 2##
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we want to show
## r(r-1) \leq a_{r+3}^2##
because this means
##\frac{1}{r(r-1)} \geq \frac{1}{a_{r+3}^2}##
noting that all values are positive for both ##(r-1)## and ##a_{r+3}##.
We can equivalently seek to prove:
## \big(r(r-1)\big)^\frac{1}{2} \leq a_{r+3}##
## \big(r(r-1)\big)^\frac{1}{2} \leq \frac{1}{2}\big(r + (r-1)\big) = r - 0.5 \leq a_{r+3}##
by ##GM \leq AM##
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Hence all we really need to prove is: ##a_{r+3} \geq r - 0.5 ##
by direct calculation, we verify this holds in the case of ##a_5## through ##a_7## as shown below.
##\begin{bmatrix}
& & a_3 & = 1.75 & \\
& & a_4 & \approx 1.6607 & \\
r = 2 & \to & a_5 & \approx 2.6663 &\gt 1.5 \\
r = 3 & \to & a_6 & \approx 3.0503 &\gt 2.5 \\
r = 4 & \to & a_7 & \approx 4.222 &\gt 3.5
\end{bmatrix}##
From here we may use induction: assume true for some ##r## now we need to prove it must be true for ##r+1##, i.e. prove
##a_{r + 4} = a_{r + 3+1} \geq (r+1) -0.5 = r+0.5 ##
using our recurrence, noting that ##\frac{1}{4 a_{r+3}a_{r+2} } \gt 0##, and then using our strong inductive hypothesis, we have
##a_{r+4} = \frac{1}{2}a_{r+3} + a_{r+2} +\frac{1}{4 a_{r+3}a_{r+2} } \gt \frac{1}{2}a_{r+3} + a_{r+2} \geq \frac{1}{2}\big(r - 0.5\big) + \big((r - 1) -0.5\big) = 1.5r - 1.75 ##
Hence we are interested in:
## 1.5r - 1.75 \geq r + 0.5##
## 0.5r \geq 2.25##
## r \geq 4.5##
Therefore the relationship must hold true for natural numbers ##r \geq 5## and we verified it holds via direct calculation for smaller values of ##r##. Thus we've proven for any natural number ##r\geq 2##
##\big(\frac{1}{a_{r+3}}\big)^2 \leq \frac{1}{r(r-1)}##
hence
##\sum_{r=2}^{n}\big(\frac{1}{a_{r+3}}\big)^2 \leq \sum_{r=2}^{n}\frac{1}{r(r-1)} \leq 1 ##