$\lim_\limits{n\to \infty}\ (\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})$? 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? 

Note that each term tends to $0$ but the number of terms tends to $\infty$.
Hint. On may write
$$
\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1}=\frac1n \cdot \sum_{k=1}^n\frac1{1+\frac{k}{n}}
$$ and recognize a Riemann sum.
