How to prove induction when n is part of the sum How to Prove
$$\sum_{k=1}^{2n} (-1)^{k-1}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{k+n}$$
by induction
we did a few of induction but the n was never a part of the sum so im kinda lost on where to start because if I would try to use n+1 i have no idea how the second sum changes

For $n=1$, LHS(=lefthand side) is 
$1 - \frac 12= \frac 12$, and 
RHS is $\frac{1}{1+1} = \frac 12$. 
Take this as the base case. 
Now do the induction step. The induction hypothesis is that  formula is true  for $n$.  We will show that this implies that it also holds for $n+1$. Let's write the LHS for $n+1$. This will be 
$$ \sum_{k=1}^{2(n+1)} (-1)^{k-1} \frac1k = \underset{(I)}{\underbrace{\sum_{k=1}^{2n} (-1)^{k-1} \frac 1k}} +\frac{1}{2n+1}- \frac{1}{2n+2}=(*)$$
By the induction hypothesis, 
$$ (I) = \sum_{k=1}^n \frac{1}{k+n}\underset{j=k-1}{=} \sum_{j=0}^{n-1} \frac{1}{j+n+1}.$$
Plugging this back into $(*)$, and isolating the $j=0$ summand, we have: 
\begin{align*} (*) &= \frac{1}{n+1} + \sum_{j=1}^{n-1} \frac{1}{j+(n+1)} + \frac{1}{n+(n+1)}-\frac{1}{(n+1)+n+1}.\\
& =\sum_{j=1}^n \frac{1}{j+(n+1)}+ \frac{1}{n+1} - \frac 12 \times\frac{1}{n+1}\\
&= \sum_{j=1}^n \frac{1}{j+(n+1)}+  \frac{1}{2(n+1)}\\
& = \sum_{j=1}^{n+1} \frac{1}{j+(n+1)},
\end{align*}
which is what we wanted to show. 
