How to prove by induction?
For $n\ge 1$: $\sum_{j=n}^{2n-1} (1/j) = \sum_{k=1}^{2n-1} ((-1)^{k+1}/k)$
1) Base case
$\sum_{j=1}^{1} (1/j) = 1 = \sum_{k=1}^{1} ((-1)^{k+1}/k)$
2) Induction [Prove that $\sum_{j=n+1}^{2(n+1)-1} (1/j) = \sum_{k=1}^{2(n+1)-1} ((-1)^{k+1}/k)$]
I can not figure out how to get from $\sum_{j=n+1}^{2(n+1)-1} (1/j)$ to $\sum_{k=1}^{2(n+1)-1} ((-1)^{k+1}/k)$.
$$\sum_{k=1}^{2(n+1)-1} ((-1)^{k+1}/k)=\sum_{k=1}^{2n-1} ((-1)^{k+1}/k)-\frac{1}{2n}+\frac{1}{2n+1}=\sum_{j=n}^{2n-1} (1/j) -\frac{1}{2n}+\frac{1}{2n+1}=( \sum_{j=n+1}^{2(n+1)-1} (1/j)+\frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+1})-\frac{1}{2n}+\frac{1}{2n+1}= \sum_{j=n+1}^{2(n+1)-1} (1/j)$$
$$ \sum_{j=n+1}^{2(n+1)-1} (1/j)=\sum_{j=n+1}^{2n+1} (1/j)= \sum_{j=n+1}^{2n-1}(1/j)+1/(2n)+1/(2n+1)= $$ $$ =\sum_{j=n}^{2n-1}(1/j)+1/(2n)+1/(2n+1)-1/n=\cdots $$
