> [!NOTE] > For all $n\in \mathbb{Z}^+,$ $\sum_{i=n+1}^{2n} \frac{1}{i} = \sum_{i=1}^{2n} (-1)^{i+1} \frac{1}{i} \tag{1}$ ###### Proof When $n=1,$ the LHS of $(1)$ equals $\frac{1}{2}.$ The RHS of $(1)$ also equals $1- \frac{1}{2}=\frac{1}{2}.$ So $(1)$ is true for $n=1.$ Suppose $(1)$ holds true for $n=k.$ Then $\begin{align} \sum_{i=k+2}^{2k+2} \frac{1}{i} &= \left( \sum_{i=k+1}^{2k} \frac{1}{i} \right)- \frac{1}{k+1} + \frac{1}{2k+1}+ \frac{1}{2(k+1)} \\ &= \sum_{i=1}^{2k+2} (-1)^{i+1} \frac{1}{i} \end{align}$which is exactly $(1)$ for $n=k+1.$ Thus by [[Induction Principle|mathematical induction]], $(1)$ holds for all $n$ as above.