Prove by mathematical induction that $forall n in mathbb{N} : sum_{k=1}^{2n} frac{(-1)^{k+1}}{k} = sum_{k=n+1}^{2n} frac{1}{k}$

Prove by mathematical induction that:

$$\forall n \in \mathbb{N} : \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$$

Step 1: Show that the statement is true for $n = 1$:

LHS = $$\frac{(-1)^{1+1}}{1} = 1$$

RHS = $$\frac{1}{1} = 1$$

Step 2: Show that "if true for n = p, then true for n = p + 1":

Starting with the LHS of equality for $n = p + 1$ and try to get to the RHS by using the equality for $n = p$. Simplifying:

$$\sum_{k=1}^{2(p+1)} \frac{(-1)^{k+1}}{k} = \sum_{k=1}^{2p+2} \frac{(-1)^{k+1}}{k}$$

Breaking out the first two term:

$$\sum_{k=1}^{2p+2} \frac{(-1)^{k+1}}{k} = \frac{(-1)^{2p+3}}{2p+2} + \frac{(-1)^{2p+2}}{2p+1} + \sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k}$$

The last term can now be exchanged for the RHS in the original equality:

$$\frac{(-1)^{2p+3}}{2p+2} + \frac{(-1)^{2p+2}}{2p+1} + \sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k} = \frac{(-1)^{2p+3}}{2p+2} + \frac{(-1)^{2p+2}}{2p+1} + \sum_{k=p+1}^{2p} \frac{1}{k}$$

Since $2p+3$ is odd and $2p+2$ is even, we get:

$$\frac{(-1)}{2p+2} + \frac{1}{2p+1} + \sum_{k=p+1}^{2p} \frac{1}{k} = \frac{(-1)}{2p+2} + \sum_{k=p+1}^{2p+1} \frac{1}{k}$$

...because we can easily absorb the second term in the third term sum. However, I am stuck here because the first term is negative. I probably made a trivial error somewhere, but I am unable to find it. Any suggestions?

Answer

When you go from $n=p$ to $n=p+1$, what you want to show is

$$\sum_{k=1}^{2(p+1)} \frac{(-1)^{k+1}}{k} = \sum_{k={\color{red}{(p+1)}}+1}^{2(p+1)} \frac{1}{k}.$$

The crucial part I think you're overlooking is in red. To get the desired result, you have to pull the first term out of your sum $\sum_{k=p+1}^{2p+1} \frac{1}{k}$.

$$\begin{align}\\ \frac{(-1)}{2p+2} + \sum_{k=p+1}^{2p+1} \frac{1}{k} &= \frac{-1}{2p+1} + \frac{1}{p+1} + \sum_{k=p+2}^{2p+1} \frac{1}{k},\\ &= \frac{-1}{2p+2} +\frac{2}{2p+2} + \sum_{k=p+2}^{2p+1} \frac{1}{k}, \\ &= \frac{1}{2p+2} + \sum_{k=p+2}^{2p+1} \frac{1}{k}, \\ &= \sum_{k=p+2}^{2p+2} \frac{1}{k}.\\ \end{align}$$