<s>[INST] this sum inequality $\sum_{i=1}^{n}\frac{1}{4i(i+1)-1}<\frac{2}{7}$ show that
$$\sum_{i=1}^{n}\dfrac{1}{4i(i+1)-1}<\dfrac{2}{7}\tag{1}$$
we have
$$4i(i+1)-1>4i^2$$
But
$$\sum_{i=1}^{n}\dfrac{1}{i^2}<\dfrac{8}{7}$$ it is clear not hold, because
$$\sum_{i=1}^{n}\dfrac{1}{i^2}>1+\dfrac{1}{4}>1+\dfrac{1}{7}=\dfrac{8}{7}$$
I can't prove that this (1) is true. Can anyone help?
 [/INST] Your estimation of denominator wasn't enough to bound the sum. I have noticed that using $4k^2 + 4k- 8$ your method will work.
\begin{align*}
\sum_{k=1}^n \frac{1}{4k^2+4k-1} & \le \frac 19+\sum_{k=2}^n \frac{1}{4k^2+4k-8} = \frac 19+\sum_{k=2}^n \left(\frac{1}{12(k-1)} - \frac{1}{12(k+2)}\right) \\
& =  \frac 19 + \frac{11}{72}- \frac{1}{12n}- \frac{1}{12n+12}- \frac{1}{12n+24} < \frac{19}{72} < \frac 27,
\end{align*}
because the sum is telescopic.
</s>