Prove $\sum_{k=1}^{2n-1} \frac{(-1)^{k + 1}}{k} = \sum_{k=n}^{2n-1} \frac{1}{k}$ by induction

Question

Question:

Prove the following statement by induction for all $n \in \mathbb{N}$:

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

My Attempt:

Base Case: For $n = 1, \sum_{k=1}^1 \frac{(-1)^{k+1}}{k} = \sum_{k=1}^1 \frac{1}{k} = 1.$

Inductive Step:

\begin{align} \sum_{k=1}^{2(n+1)-1} \frac{(-1)^{k+1}}{k} &= \sum_{k=1}^{2n+1} \frac{(-1)^{k+1}}{k} \\ &= \sum_{k=1}^{2n-1} \frac{(-1)^{k+1}}{k} + \frac{(-1)^{2n+1}}{2n} + \frac{(-1)^{2n+2}}{2n+1} \\ &= \sum_{k=n}^{2n-1} \frac{1}{k} - \frac{1}{2n} + \frac{1}{2n + 1} \end{align}

The issue here is that I'm not really sure how to go from that equation to $\sum_{k=n+1}^{2n+1} \frac{1}{k}$. I'd really appreciate it if you could guide me in the right direction.

Does this answer your question? Prove by induction that $\sum_{k=n+1}^{2n} \frac{1}{k} = \sum_{m=1}^{2n} \frac{(-1)^{m+1}}{m}, \qquad \forall n \in N$ - found using an Approach0 search. Note your problem is equivalent by replacing $n$ with $n+1$ to get ... — John Omielan, Aug 11 '23 at 02:42
(cont.) $\sum_{k=1}^{2n+1}\frac{(-1)^{k+1}}{k}=\sum_{k=n+1}^{2n+1}\frac{1}{k}$, where the last term on each side is $\frac{1}{2n+1}$, so they may be subtracted to get the expression in the proposed duplicate question. Note there are also other duplicate, or basically duplicate, questions, e.g., Summation of reciprocals, Having problem in last step on proving by induction $\sum^{2n}{i=n+1}\frac{1}{i}=\sum^{2n}{i=1}\frac{(-1)^{1+i}}{i} $ for $n\ge 1$, How do you prove this without using induction?, ... — John Omielan, Aug 11 '23 at 02:46
(cont.) Problem with proof by induction, etc. Last, but not least, welcome to Math SE. — John Omielan, Aug 11 '23 at 02:47

score 4 · Accepted Answer · answered Aug 11 '23 at 02:33

You just need to write $\sum_{k=n}^{2n-1}{\frac{1}{k}} = \sum_{k=n+1}^{2n+1}{\frac{1}{k}}+\frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+1} = \sum_{k=n+1}^{2n+1}{\frac{1}{k}}+\frac{1}{2n}-\frac{1}{2n+1}$

The last two terms will be canceled by the remaining terms in your derivation.

Prove $\sum_{k=1}^{2n-1} \frac{(-1)^{k + 1}}{k} = \sum_{k=n}^{2n-1} \frac{1}{k}$ by induction

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