I'm trying to understand an exercise on series, and I can't understand the following equality in the exercise:
$$\left|\frac{-2(n+1)}{n^2+1}\right| = 2\frac{2n+1}{n^2+1} \leq 2\frac{2}{n} = \frac{4}{n}$$ where $n \geq 0 $.
I was thinking that : $\left|-2(n+1)\right| = 2(n+1) = 2n+2 \,$, thus $\left|\frac{-2(n+1)}{n^2+1}\right| = 2\frac{n+1}{n^2+1} \leq \frac{2}{n}$
But I don't understand where the extra $2$ comes from in $2 (2n+1)$ ?
Thank you.
That must be a typo. That equality is false for $n=1$