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
 A: When you are evaluating a complex sum, it is often helpful to try a few cases.
We can have the case $n=1$, and we get
$$\sum_{k=1}^{2}(-1)^{k-1}\frac{1}{k}=\sum_{k=1}^{1}\frac{1}{k+n}$$
$$\frac{1}{1}-\frac{1}{2}=\frac{1}{2}\text{.}$$
Indeed, the equation holds.
Let's try a few more values of $n$ to see if there is a clear pattern:
$$n=2$$
$$\sum_{k=1}^{4}(-1)^{k-1}\frac{1}{k}=\sum_{k=1}^{2}\frac{1}{k+n}$$
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}=\frac{1}{3}+\frac{1}{4}$$
$$n=3$$
$$\sum_{k=1}^{6}(-1)^{k-1}\frac{1}{k}=\sum_{k=1}^{3}\frac{1}{k+n}$$
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}$$
Looking at this last equation, we see some clear patterns in change. The left side is pretty easy to see, but the right side is a little more subtle. In particular,
$$\left(\sum_{k=1}^{n}\frac{1}{k+n}\right)-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}=\sum_{k=1}^{n+1}\frac{1}{k+n+1}\text{.}$$
On the left hand side, we have that
$$\left(\sum_{k=1}^{2n}(-1)^{k-1}\frac{1}{k}\right)+\frac{1}{2n+1}-\frac{1}{2n+2}=\sum_{k=1}^{2(n+1)}(-1)^{k-1}\frac{1}{k}\text{.}$$
Formulate the inductive step, use a base case, and you are done!
A: 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. 
