$\lim_\limits{n\to \infty}\ (\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})$? what is the value given to this limit?
$$\lim_{n\to \infty}\ \bigg(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + \cdots + \frac{1}{n+1}\bigg)$$
Is it simply 0 because each term tends to 0 and you are just summing up zeros? 

By using this, 
we get $\lim_{n\to \infty}(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})=\ln{2}$.
