## Calculus, 10th Edition (Anton)

General term: $$\bigg\{ \frac{2n-1}{2n} \bigg\}_{n=1}^{+\infty}$$ The sequence converges to $2$.
As can be seen from the question, the numerators are odd numbers whereas the denominators are all even numbers. So, $n^{th}$ odd number is given by $(2n-1)$ and $n^{th}$ even number is given by $(2n)$. Thus, general term is $$\frac{2n-1}{2n}$$ The limit $$\lim_{n\to\infty}{ \frac{2n-1}{2n}}=1$$ because degree of n in both numerator and denominator is same, i.e., $1$.