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?

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 $\sum_{k=p+1}^{2p+1} \frac{1}{k}$.

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

You want to show that $\forall n \in \mathbb{N} : \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$.

Consider how each side changes when you go from $n$ to $n+1$.

The left side goes from $\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k}$ to $\sum_{k=1}^{2(n+1)} \frac{(-1)^{k+1}}{k} =\sum_{k=1}^{2n+2} \frac{(-1)^{k+1}}{k} $. This changes by $\sum_{k=2n+1}^{2n+2} \frac{(-1)^{k+1}}{k} =\frac{(-1)^{2n+2}}{2n+1}+\frac{(-1)^{2n+3}}{2n+2} =\frac{1}{2n+1}-\frac{1}{2n+2} =\frac{(2n+2)-(2n+1)}{(2n+1)(2n+2)} =\frac{1}{(2n+1)(2n+2)} $.

The right side changes from $\sum_{k=n+1}^{2n} \frac{1}{k}$ to $\sum_{k=n+2}^{2n+2} \frac{1}{k}$. The difference is

$\begin{array}\\ \sum_{k=n+2}^{2n+2} \frac{1}{k}-\sum_{k=n+1}^{2n} \frac{1}{k} &=(\sum_{k=n+2}^{2n} \frac{1}{k}+\frac1{2n+1}+\frac1{2n+2}) -(\frac1{n+1}+\sum_{k=n+1}^{2n} \frac{1}{k})\\ &=(\frac1{2n+1}+\frac1{2n+2})-(\frac1{n+1})\\ &=\frac{(2n+2)(n+1)+(2n+1)(n+1)-(2n+1)(2n+2)}{(2n+1)(2n+2)(n+1)}\\ &=\frac{(2n^2+4n+2)+(2n^2+3n+1)-(4n^2+6n+2)}{(2n+1)(2n+2)(n+1)}\\ &=\frac{n+1}{(2n+1)(2n+2)(n+1)}\\ &=\frac{1}{(2n+1)(2n+2)}\\ \end{array} $

These two are the same, so the two sums change by the same amount when $n$ goes from $n$ to $n+1$.

Notice how the key step in getting the difference for the right side was finding the terms in the two sums that have the same range of indices so they can be cancelled out. The first sum goes from $n+2$ to $2n+2$, and the second goes from $n+1$ to $2n$. What they have in common is $n+2$ to $2n$. The first sum has, in addition to the common terms, $2n+1$ and $2n+2$ (at the top); the second sum has $n+1$ (at the bottom). When the two are subtracted, only the $2n+1$, $2n+2$, and $n+1$ terms are left.

This is a very common step when doing this type of problem and you should practice to get comfortable with this.

Note: I realize after writing this that the computation of the difference of the right side could be simpler, like this:

$\begin{array}\\ \sum_{k=n+2}^{2n+2} \frac{1}{k}-\sum_{k=n+1}^{2n} \frac{1}{k} &=(\sum_{k=n+2}^{2n} \frac{1}{k}+\frac1{2n+1}+\frac1{2n+2}) -(\frac1{n+1}+\sum_{k=n+1}^{2n} \frac{1}{k})\\ &=(\frac1{2n+1}+\frac1{2n+2})-(\frac1{n+1})\\ &=\frac{(2n+2)+(2n+1)-2(2n+1)}{(2n+1)(2n+2)}\\ &=\frac{(4n+3)-(4n+2)}{(2n+1)(2n+2)}\\ &=\frac{1}{(2n+1)(2n+2)}\\ \end{array} $