<s>[INST] 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?
 [/INST] 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 $\sum_{k=p+1}^{2p+1} \frac{1}{k}$.
$$\begin{align}\\
\frac{(-1)}{2p+2} + \sum_{k=p+1}^{2p+1} \frac{1}{k} &= \frac{-1}{2p+1} + \frac{1}{p+1} + \sum_{k=p+2}^{2p+1} \frac{1}{k},\\
&= \frac{-1}{2p+2} +\frac{2}{2p+2} + \sum_{k=p+2}^{2p+1} \frac{1}{k}, \\
&= \frac{1}{2p+2} + \sum_{k=p+2}^{2p+1} \frac{1}{k}, \\
&= \sum_{k=p+2}^{2p+2} \frac{1}{k}.\\
\end{align}$$
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