<s>[INST] $\lim_\limits{n\to \infty}\ (\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})$? 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? 
 [/INST] Alternatively,
\begin{align*}
  1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n} &=
  \ln n+\gamma+\frac{1}{2n}+O\left( \frac{1}{n^2} \right) \\
  \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n} &=
  \left[ \ln (2n)+\gamma+\frac{1}{2(2n)} \right]-
  \left( \ln n+\gamma+\frac{1}{2n} \right)+
  O\left( \frac{1}{n^2} \right) \\
  &= \ln 2-\frac{1}{4n}+O\left( \frac{1}{n^2} \right)
\end{align*}
</s>