Progressions Inequality

take only 1+ 1/2+ 1/7+ 1/11 and neglect others

take? what do you mean?
I need a complete answer.

It is a very weak inequality.

\sum_{n=1}^{\infty} \frac{2}{2+n(n-1)} < 2 \left(1+ \sum_{n=2}^{\infty} \frac{1}{n(n-1)} \right) = 4< 2 \pi

I got the nth term.
But can you please tell in detail how you broke the sum in to that inequality. I am weak in this.
Thanks alot

When n>1, T_n = \frac{2}{2+n(n-1)} < \frac{2}{n(n-1)}

so this should be the inequality according to your logic,

\sum_{n=1}^{\infty}{}\frac{2}{2+n(n-1)} < 2\left(\sum_{n=2}^{\infty}{} \frac{1}

How did that extra term (1) come inside the bracket?

Sorry I don't know how to copy the equation from latex editor.